Chapter 1 Number Systems

Solution:

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The given expression is $\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}}$.

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We use the property of exponents $(a^m)^n = a^{mn}$. Here, $a = \frac{5}{6}$, $m = \frac{1}{5}$, and $n = -\frac{1}{6}$.

Simplifying the given expression:

$\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}} = \left(\frac {5}{6}\right)^{\frac{1}{5} \times \left(-\frac{1}{6}\right)}$

$\left(\frac {5}{6}\right)^{\frac{1}{5} \times \left(-\frac{1}{6}\right)} = \left(\frac {5}{6}\right)^{-\frac{1 \times 1}{5 \times 6}} = \left(\frac {5}{6}\right)^{-\frac{1}{30}}$

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Now, let's evaluate each option:

Option (A): $\left( \frac{5}{6} \right)^{\frac{1}{5}-\frac{1}{6}}$

Simplify the exponent: $\frac{1}{5} - \frac{1}{6} = \frac{6 - 5}{30} = \frac{1}{30}$.

So, option (A) is $\left( \frac{5}{6} \right)^{\frac{1}{30}}$.

This is not equal to $\left( \frac{5}{6} \right)^{-\frac{1}{30}}$.

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Option (B): $ \frac{1}{\left[ \left( \frac{5}{6} \right)^{\frac{1}{5}} \right]^\frac{1}{6}} $

Simplify the denominator using $(a^m)^n = a^{mn}$:

$\left[ \left( \frac{5}{6} \right)^{\frac{1}{5}} \right]^\frac{1}{6} = \left(\frac{5}{6}\right)^{\frac{1}{5} \times \frac{1}{6}} = \left(\frac{5}{6}\right)^{\frac{1}{30}}$

So, option (B) is $ \frac{1}{\left(\frac{5}{6}\right)^{\frac{1}{30}}} $. Using the property $\frac{1}{a^n} = a^{-n}$, this becomes $\left(\frac{5}{6}\right)^{-\frac{1}{30}}$.

This is equal to the given expression.

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Option (C): $\left(\frac{6}{5} \right)^{\frac{1}{30}}$

We know that $\frac{6}{5} = \left(\frac{5}{6}\right)^{-1}$.

Using the property $(a^m)^n = a^{mn}$:

$\left(\frac{6}{5} \right)^{\frac{1}{30}} = \left(\left(\frac{5}{6}\right)^{-1}\right)^{\frac{1}{30}} = \left(\frac{5}{6}\right)^{-1 \times \frac{1}{30}} = \left(\frac{5}{6}\right)^{-\frac{1}{30}}$

This is equal to the given expression.

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Option (D): $\left(\frac{5}{6} \right)^{-\frac{1}{30}}$

This is exactly the simplified form of the given expression.

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Comparing the simplified given expression $\left(\frac {5}{6}\right)^{-\frac{1}{30}}$ with the simplified options, we see that option (A) is $\left( \frac{5}{6} \right)^{\frac{1}{30}}$, while options (B), (C), and (D) are all equal to $\left( \frac{5}{6} \right)^{-\frac{1}{30}}$.

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Therefore, the expression which is not equal to $\left[\left(\frac {5}{6}\right)^\frac{1}{5}\right]^{-\frac{1}{6}}$ is $\left( \frac{5}{6} \right)^{\frac{1}{5}-\frac{1}{6}}$.

Solution:

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A rational number is any number that can be expressed as the quotient or fraction $\frac{p}{q}$ of two integers, where $p$ and $q$ are integers and $q \neq 0$. Examples include $\frac{1}{2}$, $-3$, $0$, $4.5$ (which is $\frac{9}{2}$).

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Let's consider the given options:

(A) Natural numbers: These are $1, 2, 3, ...$. Not all rational numbers are natural numbers (e.g., $\frac{1}{2}$ is rational but not natural).

(B) Integers: These are ..., $-3, -2, -1, 0, 1, 2, 3, ...$. Not all rational numbers are integers (e.g., $\frac{1}{2}$ is rational but not an integer).

(D) Whole numbers: These are $0, 1, 2, 3, ...$. Not all rational numbers are whole numbers (e.g., $-3$ or $\frac{1}{2}$ are rational but not whole numbers).

(C) Real numbers: The set of real numbers includes both rational numbers and irrational numbers. By definition, every rational number is a part of the set of real numbers.

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Therefore, every rational number is a real number.

Solution:

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Let $a$ and $b$ be two distinct rational numbers, with $a < b$.

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We know that the set of rational numbers is dense in the set of real numbers. This means that between any two distinct rational numbers, there exists at least one other rational number.

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Consider the average of $a$ and $b$: $\frac{a+b}{2}$. Since $a$ and $b$ are rational numbers, their sum $a+b$ is rational, and dividing by $2$ (which is a non-zero rational number) results in a rational number. Thus, $\frac{a+b}{2}$ is a rational number.

Furthermore, since $a < b$, we have $a+a < a+b$ and $a+b < b+b$. Dividing by 2 (a positive number), we get $\frac{2a}{2} < \frac{a+b}{2}$ and $\frac{a+b}{2} < \frac{2b}{2}$, which simplifies to $a < \frac{a+b}{2} < b$.

So, there is at least one rational number between $a$ and $b$.

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Now, consider the rational numbers $a$ and $\frac{a+b}{2}$. Since $a < \frac{a+b}{2}$, there must be a rational number between them, such as $\frac{a + \frac{a+b}{2}}{2} = \frac{\frac{2a+a+b}{2}}{2} = \frac{3a+b}{4}$.

We can continue this process indefinitely, finding new rational numbers between any two existing rational numbers. For example, we can find a rational number between $\frac{a+b}{2}$ and $b$, and so on.

This shows that between any two distinct rational numbers, there are infinitely many rational numbers.

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Let's evaluate the options:

(A) there is no rational number - This is false.

(B) there is exactly one rational number - This is false; we can find more than one.

(C) there are infinitely many rational numbers - This is true based on the density property.

(D) there are only rational numbers and no irrational numbers - This is false. While there are infinitely many rational numbers, there are also infinitely many irrational numbers between any two distinct rational numbers. For example, consider the irrational number $\sqrt{2}$. Between any two rational numbers $r_1$ and $r_2$, you can find rational multiples of $\sqrt{2}/n$ for a sufficiently large integer $n$, such that $r_1 < r_1 + k(\sqrt{2}/n) < r_2$ for some integer $k$. More simply, if $r_1 < r_2$, consider the interval $(r_1, r_2)$. The length of this interval is $r_2 - r_1 > 0$. Within any interval of positive length on the real number line, there are infinitely many irrational numbers.

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Based on the property of density of rational numbers, there are infinitely many rational numbers between two distinct rational numbers.

Solution:

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The decimal representation of a rational number can be of two types:

1. Terminating: The decimal representation ends after a finite number of digits. For example, $\frac{1}{4} = 0.25$, $\frac{3}{8} = 0.375$.

2. Non-terminating repeating: The decimal representation continues infinitely, but a block of digits repeats periodically. For example, $\frac{1}{3} = 0.333... = 0.\overline{3}$, $\frac{2}{7} = 0.285714285714... = 0.\overline{285714}$.

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An irrational number is a number that cannot be expressed as a simple fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The decimal representation of an irrational number is always non-terminating and non-repeating. This means the decimal digits continue infinitely without any repeating pattern. For example, $\sqrt{2} \approx 1.41421356...$, $\pi \approx 3.14159265...$.

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Let's examine the given options:

(A) terminating: A rational number can have a terminating decimal representation (e.g., $0.5 = \frac{1}{2}$).

(B) non-terminating: A rational number can have a non-terminating decimal representation (if it is repeating, e.g., $0.\overline{3}$).

(C) non-terminating repeating: A rational number can have a non-terminating repeating decimal representation (e.g., $0.\overline{3} = \frac{1}{3}$).

(D) non-terminating non-repeating: A rational number cannot have a non-terminating non-repeating decimal representation. This type of decimal representation corresponds to an irrational number.

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Therefore, the decimal representation of a rational number cannot be non-terminating non-repeating.

Solution:

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An irrational number is a number that cannot be expressed as a simple fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples include $\sqrt{2}$, $\sqrt{3}$, $\pi$, $e$.

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Let's consider some examples of the product of two irrational numbers:

Example 1: Consider the two irrational numbers $\sqrt{2}$ and $\sqrt{2}$.

Their product is $\sqrt{2} \times \sqrt{2} = \sqrt{2 \times 2} = \sqrt{4} = 2$.

The number $2$ is a rational number (it can be written as $\frac{2}{1}$).

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Example 2: Consider the two irrational numbers $\sqrt{2}$ and $\sqrt{3}$.

Their product is $\sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6}$.

The number $\sqrt{6}$ is an irrational number (as $6$ is not a perfect square).

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From these examples, we see that the product of two irrational numbers can be either a rational number (as in Example 1) or an irrational number (as in Example 2).

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Therefore, the product of any two irrational numbers is sometimes rational and sometimes irrational.

Solution:

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The number $\sqrt{2}$ is an irrational number. By definition, an irrational number cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

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The decimal expansion of any rational number is either terminating or non-terminating and repeating (recurring).

The decimal expansion of any irrational number is always non-terminating and non-repeating (non-recurring).

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Let's consider the given options:

(A) a finite decimal: This means the decimal expansion terminates. This is characteristic of some rational numbers, not irrational numbers.

(B) 1.41421: This is a finite approximation of $\sqrt{2}$. The actual decimal expansion of $\sqrt{2}$ is infinite.

(C) non-terminating recurring: "Recurring" means repeating. A non-terminating repeating decimal is characteristic of rational numbers that are not terminating decimals.

(D) non-terminating non-recurring: "Non-recurring" means non-repeating. This is the defining characteristic of the decimal expansion of an irrational number.

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Since $\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-recurring.

Solution:

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A rational number is any number that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

An irrational number is a number that cannot be expressed as a simple fraction $\frac{p}{q}$. Their decimal expansions are non-terminating and non-repeating.

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Let's evaluate each option:

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Option (A): $\sqrt{\frac{4}{9}}$

We can simplify this expression:

$\sqrt{\frac{4}{9}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$

Since $\frac{2}{3}$ is in the form $\frac{p}{q}$ where $p=2$ and $q=3$ are integers and $q \neq 0$, this is a rational number.

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Option (B): $\frac{\sqrt{12}}{\sqrt{3}}$

We can simplify this expression using the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:

$\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4}$

$\sqrt{4} = 2$

Since $2$ can be written as $\frac{2}{1}$, which is in the form $\frac{p}{q}$ where $p=2$ and $q=1$ are integers and $q \neq 0$, this is a rational number.

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Option (C): $\sqrt{7}$

The number $7$ is not a perfect square (it cannot be obtained by squaring an integer). The square root of a positive integer that is not a perfect square is an irrational number.

Thus, $\sqrt{7}$ is an irrational number.

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Option (D): $\sqrt{81}$

We can simplify this expression:

$\sqrt{81} = 9$

Since $9$ can be written as $\frac{9}{1}$, which is in the form $\frac{p}{q}$ where $p=9$ and $q=1$ are integers and $q \neq 0$, this is a rational number.

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Comparing the results, we find that only $\sqrt{7}$ is an irrational number.

Solution:

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A number is rational if its decimal expansion is either terminating or non-terminating and repeating (recurring).

A number is irrational if its decimal expansion is non-terminating and non-repeating (non-recurring).

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Let's examine the decimal expansion of each option:

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Option (A): $0.14$

This is a terminating decimal (it ends after two digits). Therefore, it represents a rational number. ($0.14 = \frac{14}{100}$)

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Option (B): $0.14\overline{16}$

This is a non-terminating repeating decimal (the block of digits '16' repeats indefinitely after the '14'). Therefore, it represents a rational number.

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Option (C): $0.\overline{1416}$

This is a non-terminating repeating decimal (the block of digits '1416' repeats indefinitely from the first decimal place). Therefore, it represents a rational number.

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Option (D): $0.4014001400014...$

This decimal expansion continues indefinitely (non-terminating). The pattern of digits '4014', '40014', '400014', etc., shows an increasing number of zeros between '4' and '14'. There is no fixed block of digits that repeats throughout the expansion. Therefore, it is a non-terminating non-repeating decimal.

This represents an irrational number.

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Based on the classification of decimal expansions, the number that is irrational is $0.4014001400014...$

Solution:

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We are asked to find a rational number that lies between $\sqrt{2}$ and $\sqrt{3}$.

First, let's consider the approximate values of $\sqrt{2}$ and $\sqrt{3}$.

$\sqrt{2} \approx 1.414$

$\sqrt{3} \approx 1.732$

So, we are looking for a rational number $x$ such that $1.414 < x < 1.732$.

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Let's examine each option:

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Option (A): $\frac{\sqrt{2}\;+\;\sqrt{3}}{2}$

This expression represents the average of $\sqrt{2}$ and $\sqrt{3}$. Since the sum of two irrational numbers ($\sqrt{2}$ and $\sqrt{3}$) is generally irrational (specifically, $\sqrt{2}+\sqrt{3}$ is irrational), and dividing by a non-zero rational number (2) results in an irrational number, $\frac{\sqrt{2}+\sqrt{3}}{2}$ is an irrational number.

Although this number lies between $\sqrt{2}$ and $\sqrt{3}$ (because if $a < b$, then $a < \frac{a+b}{2} < b$), it is not a rational number. So, option (A) is incorrect.

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Option (B): $\frac{\sqrt{2}\;.\;\sqrt{3}}{2}$

This expression simplifies to $\frac{\sqrt{6}}{2}$.

Since $6$ is not a perfect square, $\sqrt{6}$ is an irrational number. The quotient of an irrational number ($\sqrt{6}$) and a non-zero rational number (2) is an irrational number.

Let's check if it lies between $\sqrt{2}$ and $\sqrt{3}$. We can square the numbers to compare: $(\sqrt{2})^2 = 2$ $(\frac{\sqrt{6}}{2})^2 = \frac{6}{4} = 1.5$ $(\sqrt{3})^2 = 3$

Comparing the squares: $2 > 1.5$. This means $\sqrt{2} > \frac{\sqrt{6}}{2}$.

Since $\frac{\sqrt{6}}{2}$ is less than $\sqrt{2}$, it is not between $\sqrt{2}$ and $\sqrt{3}$. So, option (B) is incorrect.

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Option (C): 1.5

This number can be written as $\frac{15}{10} = \frac{3}{2}$. Since it can be expressed as a fraction of two integers with a non-zero denominator, 1.5 is a rational number.

Now let's check if it lies between $\sqrt{2}$ and $\sqrt{3}$:

We compare $1.5$ with $\sqrt{2}$ and $\sqrt{3}$ by squaring them:

$(1.5)^2 = 2.25$

$(\sqrt{2})^2 = 2$

$(\sqrt{3})^2 = 3$

Comparing the squares: $2 < 2.25 < 3$. This means $\sqrt{2} < 1.5 < \sqrt{3}$.

So, 1.5 is a rational number between $\sqrt{2}$ and $\sqrt{3}$. Option (C) is a possible answer.

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Option (D): 1.8

This number can be written as $\frac{18}{10} = \frac{9}{5}$. Since it can be expressed as a fraction of two integers with a non-zero denominator, 1.8 is a rational number.

Now let's check if it lies between $\sqrt{2}$ and $\sqrt{3}$:

Compare $1.8$ with $\sqrt{2}$ and $\sqrt{3}$ by squaring them:

$(1.8)^2 = 3.24$

$(\sqrt{2})^2 = 2$

$(\sqrt{3})^2 = 3$

Comparing the squares: $2 < 3 < 3.24$. This means $\sqrt{2} < \sqrt{3} < 1.8$.

So, 1.8 is greater than $\sqrt{3}$, and therefore not between $\sqrt{2}$ and $\sqrt{3}$. Option (D) is incorrect.

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Based on the evaluation of all options, the only rational number that lies between $\sqrt{2}$ and $\sqrt{3}$ is 1.5.

Solution:

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Let $x = 1.999...$

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Since only one digit (9) is repeating, we multiply equation (i) by $10^1 = 10$.

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Now, subtract equation (i) from equation (ii):

$\begin{array}{rc} & 10x & = & 19.999... \\ - & x & = & \phantom{0}1.999... \\ \hline & 9x & = & 18.000... \\ \hline \end{array}$

So, we have $9x = 18$.

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Divide by 9 to solve for $x$:

$x = \frac{18}{9}$

$x = 2$

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The value of 1.999... in the form $\frac{p}{q}$ is 2. We can express 2 as $\frac{2}{1}$, where $p=2$ and $q=1$ are integers and $q \neq 0$.

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Comparing this result with the given options:

(A) $\frac{19}{10} = 1.9$

(B) $\frac{1999}{1000} = 1.999$ (This is a terminating decimal, not the repeating one)

(C) 2

(D) $\frac{1}{9} = 0.111... = 0.\overline{1}$

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The correct value is 2.

Solution:

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The given expression is $2\sqrt{3}\;+\;\sqrt{3}$.

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We can think of $\sqrt{3}$ as a common factor or a variable. Let $x = \sqrt{3}$. Then the expression becomes $2x + x$.

Combining the like terms:

$2x + x = (2+1)x = 3x$

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Substituting back $x = \sqrt{3}$, we get:

$3x = 3\sqrt{3}$

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So, $2\sqrt{3}\;+\;\sqrt{3} = 3\sqrt{3}$.

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Comparing this result with the given options:

(A) $2\sqrt{6}$

(B) 6

(C) $3\sqrt{3}$

(D) $4\sqrt{6}$

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The result matches option (C).

Solution:

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The given expression is $\sqrt{10} \;\times\; \sqrt{15}$.

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We can use the property of square roots which states that for non-negative numbers $a$ and $b$, $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$.

Applying this property to the given expression:

$\sqrt{10} \;\times\; \sqrt{15} = \sqrt{10 \times 15}$

$\sqrt{10 \times 15} = \sqrt{150}$

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Now, we need to simplify $\sqrt{150}$ by finding any perfect square factors of 150.

We can find the prime factorization of 150:

$\begin{array}{c|cc} 2 & 150 \\ \hline 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So, $150 = 2 \times 3 \times 5 \times 5 = 2 \times 3 \times 5^2$.

Now, substitute this back into the square root:

$\sqrt{150} = \sqrt{2 \times 3 \times 5^2}$

Using the property $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$:

$\sqrt{2 \times 3 \times 5^2} = \sqrt{2 \times 3} \times \sqrt{5^2}$

$\sqrt{2 \times 3} \times \sqrt{5^2} = \sqrt{6} \times 5$

So, $\sqrt{150} = 5\sqrt{6}$.

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Comparing this result with the given options:

(A) $6\sqrt{5}$

(B) $5\sqrt{6}$

(C) $\sqrt{25} = 5$

(D) $10\sqrt{5}$

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The simplified value of $\sqrt{10} \;\times\; \sqrt{15}$ is $5\sqrt{6}$, which matches option (B).

Solution:

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The given expression is $\frac{1}{\sqrt{7} \;-\; 2}$.

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To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator.

The conjugate of $\sqrt{7} \;-\; 2$ is $\sqrt{7} \;+\; 2$.

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Multiply the numerator and the denominator by $\sqrt{7} \;+\; 2$:

$\frac{1}{\sqrt{7} \;-\; 2} = \frac{1}{(\sqrt{7} \;-\; 2)} \times \frac{(\sqrt{7} \;+\; 2)}{(\sqrt{7} \;+\; 2)}$

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Simplify the numerator:

$1 \times (\sqrt{7} \;+\; 2) = \sqrt{7} \;+\; 2$

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Simplify the denominator using the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$.

Here, $a = \sqrt{7}$ and $b = 2$.

$(\sqrt{7} \;-\; 2)(\sqrt{7} \;+\; 2) = (\sqrt{7})^2 \;-\; (2)^2$

$(\sqrt{7})^2 \;-\; (2)^2 = 7 \;-\; 4 = 3$

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So, the rationalized expression is:

$\frac{1 \times (\sqrt{7} \;+\; 2)}{(\sqrt{7} \;-\; 2)(\sqrt{7} \;+\; 2)} = \frac{\sqrt{7} \;+\; 2}{3}$

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Comparing the result with the given options, we find that the number obtained is $\frac{\left( \sqrt{7} \;+\; 2 \right)}{3}$.

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This matches option (A).

Solution:

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The given expression is $\frac{1}{\sqrt{9} \;-\; \sqrt{8}}$.

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First, we can simplify the terms in the denominator:

$\sqrt{9} = 3$

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$

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So, the expression becomes $\frac{1}{3 \;-\; 2\sqrt{2}}$.

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To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of $3 \;-\; 2\sqrt{2}$.

The conjugate of $3 \;-\; 2\sqrt{2}$ is $3 \;+\; 2\sqrt{2}$.

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Multiply the numerator and the denominator by $3 \;+\; 2\sqrt{2}$:

$\frac{1}{3 \;-\; 2\sqrt{2}} = \frac{1}{(3 \;-\; 2\sqrt{2})} \times \frac{(3 \;+\; 2\sqrt{2})}{(3 \;+\; 2\sqrt{2})}$

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Simplify the numerator:

$1 \times (3 \;+\; 2\sqrt{2}) = 3 \;+\; 2\sqrt{2}$

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Simplify the denominator using the difference of squares formula: $(a-b)(a+b) = a^2 - b^2$.

Here, $a = 3$ and $b = 2\sqrt{2}$.

$(3 \;-\; 2\sqrt{2})(3 \;+\; 2\sqrt{2}) = (3)^2 \;-\; (2\sqrt{2})^2$

$(3)^2 = 9$

$(2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8$

So, $(3 \;-\; 2\sqrt{2})(3 \;+\; 2\sqrt{2}) = 9 \;-\; 8 = 1$

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Thus, the expression becomes:

$\frac{3 \;+\; 2\sqrt{2}}{1} = 3 \;+\; 2\sqrt{2}$

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Comparing this result with the given options:

(A) $\frac{1}{2} \left( 3 - 2\sqrt{2} \right)$

(B) $\frac{1}{3 \;+\; 2\sqrt{2}}$

(C) $3 - 2\sqrt{2}$

(D) $3 + 2\sqrt{2}$

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The simplified and rationalized expression matches option (D).

Solution:

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The given expression is $\frac{7}{3\sqrt{3} \;-\; 2\sqrt{2}}$.

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To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator.

The denominator is $3\sqrt{3} \;-\; 2\sqrt{2}$. Its conjugate is $3\sqrt{3} \;+\; 2\sqrt{2}$.

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The rationalized expression is obtained by multiplying the fraction by $\frac{3\sqrt{3} \;+\; 2\sqrt{2}}{3\sqrt{3} \;+\; 2\sqrt{2}}$:

$\frac{7}{3\sqrt{3} \;-\; 2\sqrt{2}} = \frac{7}{(3\sqrt{3} \;-\; 2\sqrt{2})} \times \frac{(3\sqrt{3} \;+\; 2\sqrt{2})}{(3\sqrt{3} \;+\; 2\sqrt{2})}$

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We are interested in the denominator of the resulting expression. The new denominator is the product of the original denominator and its conjugate:

New Denominator $= (3\sqrt{3} \;-\; 2\sqrt{2})(3\sqrt{3} \;+\; 2\sqrt{2})$

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Using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$, where $a = 3\sqrt{3}$ and $b = 2\sqrt{2}$:

New Denominator $= (3\sqrt{3})^2 \;-\; (2\sqrt{2})^2$

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Calculate $(3\sqrt{3})^2$ and $(2\sqrt{2})^2$:

$(3\sqrt{3})^2 = 3^2 \times (\sqrt{3})^2 = 9 \times 3 = 27$

$(2\sqrt{2})^2 = 2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8$

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Substitute these values back into the denominator expression:

New Denominator $= 27 \;-\; 8 = 19$

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The denominator after rationalizing the expression is 19.

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Comparing this result with the given options:

(A) 13

(B) 19

(C) 5

(D) 35

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The correct denominator is 19.

Solution:

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The given expression is $\frac{\sqrt{32} \;+\; \sqrt{48}}{\sqrt{8} \;+\; \sqrt{12}}$.

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First, simplify each square root by factoring out perfect squares:

$\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}$

$\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3}$

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$

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Substitute the simplified square roots back into the expression:

$\frac{\sqrt{32} \;+\; \sqrt{48}}{\sqrt{8} \;+\; \sqrt{12}} = \frac{4\sqrt{2} \;+\; 4\sqrt{3}}{2\sqrt{2} \;+\; 2\sqrt{3}}$

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Now, factor out common terms from the numerator and the denominator.

Numerator: $4\sqrt{2} \;+\; 4\sqrt{3} = 4(\sqrt{2} \;+\; \sqrt{3})$

Denominator: $2\sqrt{2} \;+\; 2\sqrt{3} = 2(\sqrt{2} \;+\; \sqrt{3})$

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Substitute the factored expressions back into the fraction:

$\frac{4(\sqrt{2} \;+\; \sqrt{3})}{2(\sqrt{2} \;+\; \sqrt{3})}$

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Cancel out the common factor $(\sqrt{2} \;+\; \sqrt{3})$ from the numerator and the denominator (assuming $\sqrt{2} + \sqrt{3} \neq 0$, which is true):

$\frac{\cancel{4}(\cancel{\sqrt{2} \;+\; \sqrt{3}})}{\cancel{2}(\cancel{\sqrt{2} \;+\; \sqrt{3}})} = \frac{4}{2}$

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Simplify the fraction:

$\frac{4}{2} = 2$

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The value of the expression is 2.

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Comparing the result with the given options:

(A) $\sqrt{2}$

(B) 2

(C) 4

(D) 8

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The correct value is 2, which matches option (B).

Solution:

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The given expression is $\sqrt{\frac{\sqrt{2} \;-\; 1}{\sqrt{2} \;+\; 1}}$.

We are given that $\sqrt{2} \approx 1.4142$.

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First, let's simplify the expression inside the square root by rationalizing the denominator:

$\frac{\sqrt{2} \;-\; 1}{\sqrt{2} \;+\; 1}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2} \;-\; 1$:

$\frac{\sqrt{2} \;-\; 1}{\sqrt{2} \;+\; 1} = \frac{(\sqrt{2} \;-\; 1)}{(\sqrt{2} \;+\; 1)} \times \frac{(\sqrt{2} \;-\; 1)}{(\sqrt{2} \;-\; 1)}$

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Simplify the numerator using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$(\sqrt{2} \;-\; 1)(\sqrt{2} \;-\; 1) = (\sqrt{2} \;-\; 1)^2 = (\sqrt{2})^2 - 2(\sqrt{2})(1) + (1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$

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Simplify the denominator using the formula $(a+b)(a-b) = a^2 - b^2$:

$(\sqrt{2} \;+\; 1)(\sqrt{2} \;-\; 1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$

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So, the expression inside the square root simplifies to:

$\frac{3 - 2\sqrt{2}}{1} = 3 - 2\sqrt{2}$

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Now, the original expression becomes $\sqrt{3 - 2\sqrt{2}}$.

We need to simplify $\sqrt{3 - 2\sqrt{2}}$. This expression is of the form $\sqrt{a - \sqrt{b}}$. We look for two numbers whose sum is 3 and whose product is such that when multiplied by 4 gives $4 \times (\sqrt{2})^2 = 4 \times 2 = 8$. Alternatively, we can try to express $3 - 2\sqrt{2}$ as a perfect square $(x-y)^2 = x^2 + y^2 - 2xy$. We need $x^2 + y^2 = 3$ and $2xy = 2\sqrt{2}$, which means $xy = \sqrt{2}$. By inspection, we can see that if $x = \sqrt{2}$ and $y = 1$, then $x^2 = (\sqrt{2})^2 = 2$, $y^2 = 1^2 = 1$. $x^2 + y^2 = 2 + 1 = 3$. $2xy = 2(\sqrt{2})(1) = 2\sqrt{2}$. So, $3 - 2\sqrt{2} = (\sqrt{2})^2 - 2(\sqrt{2})(1) + 1^2 = (\sqrt{2} - 1)^2$.

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Thus, $\sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2}$.

Since $\sqrt{2} \approx 1.4142 > 1$, the value $(\sqrt{2} - 1)$ is positive. So, $\sqrt{(\sqrt{2} - 1)^2} = |\sqrt{2} - 1| = \sqrt{2} - 1$.

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Now, substitute the given value of $\sqrt{2} \approx 1.4142$ into the simplified expression $\sqrt{2} - 1$:

$\sqrt{2} - 1 \approx 1.4142 - 1 = 0.4142$

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Comparing this result with the given options:

(A) 2.4142

(B) 5.8282

(C) 0.4142

(D) 0.1718

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The calculated value 0.4142 matches option (C).

Solution:

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The given expression is $\sqrt [4] {\sqrt [3] {2^2}}$.

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We can rewrite roots using fractional exponents: $\sqrt[n]{a} = a^{1/n}$.

Also, we use the property of exponents: $(a^m)^n = a^{mn}$.

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Let's start from the innermost part of the expression:

$\sqrt [3] {2^2}$ can be written as $(2^2)^{1/3}$.

Using the property $(a^m)^n = a^{mn}$, we get:

$(2^2)^{1/3} = 2^{2 \times \frac{1}{3}} = 2^{\frac{2}{3}}$

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Now, substitute this back into the original expression:

$\sqrt [4] {\sqrt [3] {2^2}} = \sqrt [4] {2^{\frac{2}{3}}}$

Again, rewrite the outer root using a fractional exponent:

$\sqrt [4] {2^{\frac{2}{3}}} = (2^{\frac{2}{3}})^{1/4}$

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Using the property $(a^m)^n = a^{mn}$ again:

$(2^{\frac{2}{3}})^{1/4} = 2^{\frac{2}{3} \times \frac{1}{4}}$

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Simplify the exponent:

$\frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$

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So, the simplified expression is $2^{\frac{1}{6}}$.

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Comparing this result with the given options:

(A) $2^{-\frac{1}{6}}$

(B) $2^{-6}$

(C) $2^{\frac{1}{6}}$

(D) $2^{6}$

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The result matches option (C).

Solution:

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The given expression is $\sqrt [3] {2} \;.\; \sqrt [4] {2} \;.\; \sqrt [12] {32}$.

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We can rewrite the roots using fractional exponents. Recall that $\sqrt[n]{a} = a^{1/n}$.

$\sqrt [3] {2} = 2^{1/3}$

$\sqrt [4] {2} = 2^{1/4}$

For $\sqrt [12] {32}$, first express 32 as a power of 2:

$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5$.

So, $\sqrt [12] {32} = \sqrt [12] {2^5}$.

Using the fractional exponent form: $\sqrt [12] {2^5} = (2^5)^{1/12}$.

Using the property $(a^m)^n = a^{mn}$: $(2^5)^{1/12} = 2^{5 \times \frac{1}{12}} = 2^{5/12}$.

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Now, substitute these fractional exponent forms back into the product:

$\sqrt [3] {2} \;.\; \sqrt [4] {2} \;.\; \sqrt [12] {32} = 2^{1/3} \times 2^{1/4} \times 2^{5/12}$

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When multiplying powers with the same base, we add the exponents. Using the property $a^m \times a^n \times a^p = a^{m+n+p}$:

The sum of the exponents is $\frac{1}{3} + \frac{1}{4} + \frac{5}{12}$.

To add these fractions, find a common denominator, which is 12.

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

The sum of exponents is $\frac{4}{12} + \frac{3}{12} + \frac{5}{12} = \frac{4+3+5}{12} = \frac{12}{12} = 1$.

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So, the expression simplifies to $2^1$.

$2^1 = 2$.

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The value of the product is 2.

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Comparing the result with the given options:

(A) $\sqrt{2} = 2^{1/2}$

(B) 2

(C) $\sqrt [12] {2} = 2^{1/12}$

(D) $\sqrt [12] {32} = 2^{5/12}$

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The result matches option (B).

Solution:

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The given expression is $\sqrt [4] {(81)^{-2}}$.

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We can rewrite the fourth root using a fractional exponent: $\sqrt[n]{a} = a^{1/n}$.

So, $\sqrt [4] {(81)^{-2}} = ((81)^{-2})^{1/4}$.

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Using the property of exponents $(a^m)^n = a^{mn}$, we multiply the exponents:

$((81)^{-2})^{1/4} = (81)^{-2 \times \frac{1}{4}} = (81)^{-\frac{2}{4}} = (81)^{-\frac{1}{2}}$

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Now, express the base 81 as a power of 3:

$81 = 3 \times 3 \times 3 \times 3 = 3^4$

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Substitute $81 = 3^4$ into the expression:

$(81)^{-\frac{1}{2}} = (3^4)^{-\frac{1}{2}}$

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Using the property $(a^m)^n = a^{mn}$ again:

$(3^4)^{-\frac{1}{2}} = 3^{4 \times (-\frac{1}{2})} = 3^{-\frac{4}{2}} = 3^{-2}$

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Using the property of negative exponents $a^{-n} = \frac{1}{a^n}$:

$3^{-2} = \frac{1}{3^2}$

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Calculate $3^2$:

$\frac{1}{3^2} = \frac{1}{9}$

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So, the value of $\sqrt [4] {(81)^{-2}}$ is $\frac{1}{9}$.

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Comparing this result with the given options:

(A) $\frac{1}{9}$

(B) $\frac{1}{3}$

(C) 9

(D) $\frac{1}{81}$

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The result matches option (A).

Solution:

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The given expression is $(256)^{0.16} \times (256)^{0.09}$.

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We use the property of exponents for multiplication with the same base: $a^m \times a^n = a^{m+n}$.

Here, the base is $a = 256$, and the exponents are $m = 0.16$ and $n = 0.09$.

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Apply the property by adding the exponents:

$(256)^{0.16} \times (256)^{0.09} = (256)^{0.16 + 0.09}$

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Calculate the sum of the exponents:

$0.16 + 0.09 = 0.25$

So, the expression becomes $(256)^{0.25}$.

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Rewrite the decimal exponent as a fraction:

$0.25 = \frac{25}{100} = \frac{1}{4}$

So, the expression is $(256)^{\frac{1}{4}}$.

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The exponent $\frac{1}{4}$ represents the fourth root. We need to find the number which, when raised to the power of 4, equals 256. Let this number be $x$. Then $x^4 = 256$.

We can find the prime factorization of 256:

$\begin{array}{c|cc} 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$

So, $256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8$.

Alternatively, $256 = (16)^2 = (4^2)^2 = 4^4$.

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Substitute $256 = 4^4$ into the expression $(256)^{\frac{1}{4}}$:

$(4^4)^{\frac{1}{4}}$

Using the property $(a^m)^n = a^{mn}$:

$(4^4)^{\frac{1}{4}} = 4^{4 \times \frac{1}{4}} = 4^1 = 4$

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The value of the expression is 4.

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Comparing this result with the given options:

(A) 4

(B) 16

(C) 64

(D) 256.25

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The result matches option (A).

Solution:

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We need to find which of the given expressions is equal to $x$. We will simplify each option using the properties of exponents and roots.

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Recall the properties of exponents: $(a^m)^n = a^{mn}$ $a^m \times a^n = a^{m+n}$ $\sqrt[n]{a^m} = (a^m)^{1/n} = a^{m/n}$ $\sqrt{a} = a^{1/2}$

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Option (A): $x^{\frac{12}{7}} - x^{\frac{5}{7}}$

This is a subtraction of two terms. There is no general rule to simplify the difference of powers with the same base to a single power of that base. This expression is not equal to $x$.

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Option (B): $\sqrt [12] {(x^4)^{\frac{1}{3}}}$

Simplify the innermost expression first:

$(x^4)^{\frac{1}{3}} = x^{4 \times \frac{1}{3}} = x^{\frac{4}{3}}$

Now apply the outer root:

$\sqrt [12] {x^{\frac{4}{3}}} = (x^{\frac{4}{3}})^{\frac{1}{12}}$

Using the property $(a^m)^n = a^{mn}$:

$(x^{\frac{4}{3}})^{\frac{1}{12}} = x^{\frac{4}{3} \times \frac{1}{12}} = x^{\frac{4}{36}} = x^{\frac{1}{9}}$

This is not equal to $x$.

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Option (C): $(\sqrt{x^3})^{\frac{2}{3}}$

Rewrite the square root using a fractional exponent:

$\sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^{3 \times \frac{1}{2}} = x^{\frac{3}{2}}$

Now apply the outer exponent:

$(x^{\frac{3}{2}})^{\frac{2}{3}}$

Using the property $(a^m)^n = a^{mn}$:

$(x^{\frac{3}{2}})^{\frac{2}{3}} = x^{\frac{3}{2} \times \frac{2}{3}} = x^{\frac{6}{6}} = x^1 = x$

This expression simplifies to $x$.

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Option (D): $x^{\frac{12}{7}} \times x^{\frac{7}{12}}$

Using the property $a^m \times a^n = a^{m+n}$, add the exponents:

$x^{\frac{12}{7}} \times x^{\frac{7}{12}} = x^{\frac{12}{7} + \frac{7}{12}}$

Add the fractions in the exponent:

$\frac{12}{7} + \frac{7}{12} = \frac{12 \times 12}{7 \times 12} + \frac{7 \times 7}{12 \times 7} = \frac{144}{84} + \frac{49}{84} = \frac{144 + 49}{84} = \frac{193}{84}$

So, $x^{\frac{12}{7}} \times x^{\frac{7}{12}} = x^{\frac{193}{84}}$

This is not equal to $x$ (which is $x^1$).

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Based on the simplification of each option, only option (C) is equal to $x$.

Solution:

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The question asks if there exist two irrational numbers whose sum and product are both rational numbers.

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Yes, there are two such irrational numbers.

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Let us consider two numbers of the form $(a + \sqrt{b})$ and $(a - \sqrt{b})$, where $a$ is a rational number and $\sqrt{b}$ is an irrational number (i.e., $b$ is a positive number that is not a perfect square of a rational number).

For these numbers to be irrational, we need $a + \sqrt{b}$ and $a - \sqrt{b}$ to be irrational. This is true if $\sqrt{b}$ is irrational and $a$ is rational (unless $b=0$, which would make $\sqrt{b}=0$ rational; we assume $b>0$ and not a perfect square).

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Let the two numbers be $x = a + \sqrt{b}$ and $y = a - \sqrt{b}$.

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Let's find their sum:

$x + y = (a + \sqrt{b}) + (a - \sqrt{b})$

$x + y = a + \sqrt{b} + a - \sqrt{b}$

$x + y = (a + a) + (\sqrt{b} - \sqrt{b})$

$x + y = 2a + 0$

$x + y = 2a$

Since $a$ is a rational number, $2a$ is also a rational number.

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Now, let's find their product:

$x \times y = (a + \sqrt{b}) \times (a - \sqrt{b})$

Using the difference of squares formula $(A+B)(A-B) = A^2 - B^2$, where $A=a$ and $B=\sqrt{b}$:

$x \times y = a^2 - (\sqrt{b})^2$

$x \times y = a^2 - b$

Since $a$ is a rational number, $a^2$ is also a rational number. Since $b$ is the number under the square root $\sqrt{b}$, if $\sqrt{b}$ is irrational, $b$ is a positive non-perfect square number. However, the product $a^2 - b$ depends on whether $b$ itself is rational. If we choose $a$ to be rational and $\sqrt{b}$ to be irrational, then $b$ must be rational (e.g., if $\sqrt{b}=\sqrt{2}$, $b=2$; if $\sqrt{b}=\sqrt{3}$, $b=3$). So, $a^2$ is rational and $b$ is rational. The difference of two rational numbers ($a^2 - b$) is a rational number.

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Justification with an example:

Let $a=1$ (a rational number) and $\sqrt{b}=\sqrt{2}$ (an irrational number). Then $b=2$ ( a rational number).

The two numbers are $x = 1 + \sqrt{2}$ and $y = 1 - \sqrt{2}$.

Both $1+\sqrt{2}$ and $1-\sqrt{2}$ are irrational numbers (sum/difference of a non-zero rational and an irrational is irrational).

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Their sum:

$(1 + \sqrt{2}) + (1 - \sqrt{2}) = 1 + \sqrt{2} + 1 - \sqrt{2} = 1 + 1 = 2$

The sum is 2, which is a rational number.

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Their product:

$(1 + \sqrt{2}) \times (1 - \sqrt{2})$

Using $(A+B)(A-B) = A^2 - B^2$:

$(1 + \sqrt{2})(1 - \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1$

The product is -1, which is a rational number.

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Since we have found two irrational numbers ($1+\sqrt{2}$ and $1-\sqrt{2}$) whose sum (2) and product (-1) are both rational, the answer is yes.

Solution:

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The statement is: There is a number $x$ such that $x^2$ is irrational but $x^4$ is rational.

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This statement is True.

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To justify this, we need to find such a number $x$.

We are given the conditions: 1. $x^2$ is an irrational number. 2. $x^4$ is a rational number.

Note that $x^4 = (x^2)^2$. So, the second condition means that the square of the number $x^2$ is rational.

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We are looking for an irrational number ($x^2$) whose square is rational. Consider a number of the form $\sqrt{k}$, where $k$ is a positive rational number that is not a perfect square of a rational number. Such a number $\sqrt{k}$ is irrational.

If we take the square of $\sqrt{k}$, we get $(\sqrt{k})^2 = k$. Since $k$ is a rational number, $(\sqrt{k})^2$ is rational.

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Let's choose $k = 2$. The number $\sqrt{2}$ is irrational. Its square is $(\sqrt{2})^2 = 2$, which is rational.

According to our conditions, $x^2$ should be this irrational number. So, let $x^2 = \sqrt{2}$.

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Now, let's find $x$. If $x^2 = \sqrt{2}$, then $x = \sqrt{\sqrt{2}}$. We can write $\sqrt{\sqrt{2}}$ using exponents as $(2^{1/2})^{1/2} = 2^{(1/2) \times (1/2)} = 2^{1/4}$.

So, let's take $x = 2^{1/4}$ (or $x = \sqrt[4]{2}$).

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Let's check the conditions for this value of $x$:

Condition 1: $x^2$ must be irrational.

$x^2 = (2^{1/4})^2 = 2^{1/4 \times 2} = 2^{2/4} = 2^{1/2} = \sqrt{2}$.

Since 2 is not a perfect square, $\sqrt{2}$ is an irrational number. Condition 1 is satisfied.

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Condition 2: $x^4$ must be rational.

$x^4 = (2^{1/4})^4 = 2^{1/4 \times 4} = 2^1 = 2$.

The number 2 can be written as $\frac{2}{1}$, which is a rational number. Condition 2 is satisfied.

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Since we found a number $x = \sqrt[4]{2}$ such that $x^2 = \sqrt{2}$ (irrational) and $x^4 = 2$ (rational), the statement is true.

Solution:

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The question asks if the sum of a rational number $x$ and an irrational number $y$ is necessarily an irrational number.

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The statement is: Yes, the sum $x + y$ is necessarily an irrational number.

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Justification:

Let $x$ be a rational number and $y$ be an irrational number.

A rational number can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

An irrational number cannot be written in this form.

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Assume, for the sake of contradiction, that the sum $x + y$ is a rational number. Let $x + y = r$, where $r$ is a rational number.

Since $x$ is rational, we can write $x = \frac{p}{q}$ for some integers $p, q$ with $q \neq 0$.

Since $r$ is rational, we can write $r = \frac{m}{n}$ for some integers $m, n$ with $n \neq 0$.

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From our assumption $x + y = r$, we can write $y = r - x$.

Substitute the fractional forms of $r$ and $x$:

$y = \frac{m}{n} - \frac{p}{q}$

To subtract these fractions, we find a common denominator, which is $nq$ (since $n \neq 0$ and $q \neq 0$, $nq \neq 0$).

$y = \frac{m \times q}{n \times q} - \frac{p \times n}{q \times n}$

$y = \frac{mq - pn}{nq}$

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Since $m, q, p, n$ are all integers, the numerator $(mq - pn)$ is an integer, and the denominator $(nq)$ is a non-zero integer.

Therefore, $\frac{mq - pn}{nq}$ is a rational number.

This means that $y$ is a rational number.

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However, we were given that $y$ is an irrational number. This contradicts our conclusion that $y$ is rational.

The assumption that $x + y$ is rational must be false.

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Therefore, the sum of a rational number and an irrational number must be an irrational number.

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Example:

Let $x = 2$ be a rational number (since $2 = \frac{2}{1}$).

Let $y = \sqrt{3}$ be an irrational number.

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Their sum is $x + y = 2 + \sqrt{3}$.

We know that the sum of a non-zero rational number and an irrational number is irrational.

To confirm that $2 + \sqrt{3}$ is irrational, assume it is rational. Let $2 + \sqrt{3} = r$, where $r$ is rational.

Then $\sqrt{3} = r - 2$.

Since $r$ is rational and 2 is rational, their difference $r - 2$ is rational. This means $\sqrt{3}$ is rational.

However, $\sqrt{3}$ is known to be irrational. This is a contradiction.

Therefore, $2 + \sqrt{3}$ is an irrational number.

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The example supports the statement that the sum of a rational and an irrational number is necessarily irrational.

Solution:

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The question asks whether the product of a rational number $x$ and an irrational number $y$ is necessarily an irrational number.

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The statement is: No, the product $xy$ is not necessarily an irrational number.

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Justification by an example:

Let $x$ be a rational number and $y$ be an irrational number.

Consider the case where the rational number $x$ is 0.

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Let $x = 0$.

The number 0 is a rational number because it can be written as $\frac{0}{1}$, where 0 and 1 are integers and the denominator 1 is not zero.

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Let $y = \sqrt{2}$.

The number $\sqrt{2}$ is an irrational number because its decimal expansion is non-terminating and non-repeating, and it cannot be expressed as a fraction of two integers.

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Now, let's find the product of $x$ and $y$:

$x \times y = 0 \times \sqrt{2}$

$x \times y = 0$

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The product is 0. As established earlier, 0 is a rational number.

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In this example, we have a rational number ($x=0$) and an irrational number ($y=\sqrt{2}$) whose product ($0$) is a rational number.

Therefore, the product of a rational number and an irrational number is not necessarily an irrational number. It can be rational (specifically, when the rational number is 0).

Solution:

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Let's evaluate each statement and provide justification.

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(i) $\frac{\sqrt{2}}{3}$ is a rational number.

This statement is False.

Justification: $\sqrt{2}$ is an irrational number, and 3 is a non-zero rational number. The quotient of an irrational number and a non-zero rational number is always an irrational number. Therefore, $\frac{\sqrt{2}}{3}$ is an irrational number.

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(ii) There are infinitely many integers between any two integers.

This statement is False.

Justification: The integers are ... , -2, -1, 0, 1, 2, ... . Between any two distinct integers, there is a finite number of integers. For example, between 2 and 5, the integers are 3 and 4 (a finite number). Between any two consecutive integers, such as 2 and 3, there are no integers at all.

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(iii) Number of rational numbers between 15 and 18 is finite.

This statement is False.

Justification: 15 and 18 are both rational numbers. A property of rational numbers is that they are dense in the real numbers, meaning that between any two distinct rational numbers, there are infinitely many other rational numbers. For instance, if $r_1$ and $r_2$ are two rational numbers, then $\frac{r_1+r_2}{2}$ is a rational number between them. We can repeat this process indefinitely, generating infinitely many rational numbers between $r_1$ and $r_2$.

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(iv) There are numbers which cannot be written in the form $\frac{p}{q}$, q ≠ 0 , p, q both are integers.

This statement is True.

Justification: Numbers that cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$, are called irrational numbers. Examples of such numbers include $\sqrt{2}$, $\sqrt{3}$, $\pi$, and $e$. These numbers have non-terminating, non-repeating decimal expansions and thus cannot be expressed as a fraction of two integers.

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(v) The square of an irrational number is always rational.

This statement is False.

Justification: Consider the irrational number $\sqrt{2} + 1$. The sum of an irrational number ($\sqrt{2}$) and a rational number (1) is irrational, so $\sqrt{2} + 1$ is irrational.

Now, let's find the square of $\sqrt{2} + 1$:

$(\sqrt{2} + 1)^2 = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2$

$(\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1$

$(\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$

The number $3 + 2\sqrt{2}$ is the sum of a rational number (3) and an irrational number ($2\sqrt{2}$). The sum of a rational and a non-zero irrational number is irrational. Therefore, $(\sqrt{2} + 1)^2$ is irrational. This is a counterexample to the statement.

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(vi) $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.

This statement is False.

Justification: Let's simplify the expression $\frac{\sqrt{12}}{\sqrt{3}}$:

$\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4}$

$\sqrt{4} = 2$

The value of the expression is 2. The number 2 is a rational number because it can be written as $\frac{2}{1}$, where 2 and 1 are integers and $1 \neq 0$. The fact that $\sqrt{12}$ and $\sqrt{3}$ are not integers does not mean their quotient is not rational; the rationality of the quotient depends on the simplified value of the quotient itself.

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(vii) $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form p/q , q ≠ 0 and so it is a rational number.

This statement is False.

Justification: While the expression is presented as a fraction, for a number to be rational when written in the form $\frac{p}{q}$, it is required that both $p$ and $q$ must be integers and $q \neq 0$.

Let's simplify the given expression:

$\frac{\sqrt{15}}{\sqrt{3}} = \sqrt{\frac{15}{3}} = \sqrt{5}$

The number $\sqrt{5}$ is an irrational number because 5 is not a perfect square, and its decimal expansion is non-terminating and non-repeating. Although the original expression looks like a fraction, the numerator ($\sqrt{15}$) and the denominator ($\sqrt{3}$) are not integers. After simplification, the number is $\sqrt{5}$, which is irrational. Therefore, the original expression represents an irrational number.

Solution:

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We need to classify each given number as rational or irrational and provide justification.

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(i) $\sqrt{196}$

$\sqrt{196} = 14$.

Since 14 is an integer, it can be written as $\frac{14}{1}$, which is in the form $\frac{p}{q}$ where $p=14$ and $q=1$ are integers and $q \neq 0$.

Therefore, $\sqrt{196}$ is a rational number.

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(ii) $3\sqrt{18}$

Simplify the expression: $3\sqrt{18} = 3 \times \sqrt{9 \times 2} = 3 \times \sqrt{9} \times \sqrt{2} = 3 \times 3 \times \sqrt{2} = 9\sqrt{2}$.

Since $\sqrt{2}$ is an irrational number and 9 is a non-zero rational number, their product $9\sqrt{2}$ is irrational.

Therefore, $3\sqrt{18}$ is an irrational number.

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(iii) $\sqrt{\frac{9}{27}}$

Simplify the expression: $\sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

To rationalize the denominator, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

Since $\sqrt{3}$ is an irrational number and 3 is a non-zero rational number, their quotient $\frac{\sqrt{3}}{3}$ is irrational.

Therefore, $\sqrt{\frac{9}{27}}$ is an irrational number.

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(iv) $\frac{\sqrt{28}}{\sqrt{343}}$

Simplify the expression: $\frac{\sqrt{28}}{\sqrt{343}} = \sqrt{\frac{28}{343}}$.

Simplify the fraction inside the square root: $\frac{28}{343} = \frac{4 \times 7}{49 \times 7} = \frac{4}{49}$.

So, $\sqrt{\frac{28}{343}} = \sqrt{\frac{4}{49}} = \frac{\sqrt{4}}{\sqrt{49}} = \frac{2}{7}$.

Since $\frac{2}{7}$ is in the form $\frac{p}{q}$ where $p=2$ and $q=7$ are integers and $q \neq 0$, it is rational.

Therefore, $\frac{\sqrt{28}}{\sqrt{343}}$ is a rational number.

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(v) $-\sqrt{0.4}$

Rewrite the decimal as a fraction: $0.4 = \frac{4}{10} = \frac{2}{5}$.

The expression is $-\sqrt{\frac{2}{5}} = -\frac{\sqrt{2}}{\sqrt{5}}$. This is the negative of the quotient of two irrational numbers $\sqrt{2}$ and $\sqrt{5}$. The quotient $\frac{\sqrt{2}}{\sqrt{5}} = \sqrt{\frac{2}{5}} = \sqrt{0.4}$. Since 0.4 is not a perfect square, $\sqrt{0.4}$ is irrational. The negative of an irrational number is irrational.

Alternatively, $-\frac{\sqrt{2}}{\sqrt{5}} = -\frac{\sqrt{2} \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = -\frac{\sqrt{10}}{5}$. Since $\sqrt{10}$ is irrational and 5 is a non-zero rational, $-\frac{\sqrt{10}}{5}$ is irrational.

Therefore, $-\sqrt{0.4}$ is an irrational number.

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(vi) $\frac{\sqrt{12}}{\sqrt{75}}$

Simplify the expression: $\frac{\sqrt{12}}{\sqrt{75}} = \sqrt{\frac{12}{75}}$.

Simplify the fraction inside the square root: $\frac{12}{75} = \frac{3 \times 4}{3 \times 25} = \frac{4}{25}$.

So, $\sqrt{\frac{12}{75}} = \sqrt{\frac{4}{25}} = \frac{\sqrt{4}}{\sqrt{25}} = \frac{2}{5}$.

Since $\frac{2}{5}$ is in the form $\frac{p}{q}$ where $p=2$ and $q=5$ are integers and $q \neq 0$, it is rational.

Therefore, $\frac{\sqrt{12}}{\sqrt{75}}$ is a rational number.

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(vii) 0.5918

This is a terminating decimal. Terminating decimals can always be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

$0.5918 = \frac{5918}{10000}$. Since 5918 and 10000 are integers and $10000 \neq 0$, it is rational.

Therefore, 0.5918 is a rational number.

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(viii) $(1 + \sqrt{5}) - (4 + \sqrt{5})$

Simplify the expression: $(1 + \sqrt{5}) - (4 + \sqrt{5}) = 1 + \sqrt{5} - 4 - \sqrt{5}$.

Combine like terms: $(1 - 4) + (\sqrt{5} - \sqrt{5}) = -3 + 0 = -3$.

Since -3 is an integer, it can be written as $\frac{-3}{1}$, which is in the form $\frac{p}{q}$ where $p=-3$ and $q=1$ are integers and $q \neq 0$.

Therefore, $(1 + \sqrt{5}) - (4 + \sqrt{5})$ is a rational number.

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(ix) 10.124124 ...

This decimal can be written as $10.\overline{124}$. It is a non-terminating decimal with a repeating block of digits '124'.

Any non-terminating repeating decimal represents a rational number because it can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.

Therefore, 10.124124 ... is a rational number.

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(x) 1.010010001…

This is a non-terminating decimal. The pattern of digits is 01, then 001, then 0001, and so on, with an increasing number of zeros between the ones. There is no fixed block of digits that repeats periodically.

This is a non-terminating, non-repeating decimal. Numbers with non-terminating, non-repeating decimal expansions are irrational by definition.

Therefore, 1.010010001… is an irrational number.

Solution:

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To locate $\sqrt{13}$ on the number line, we can use the concept of the Pythagorean theorem.

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse ($c$) is equal to the sum of the squares of the other two sides ($a$ and $b$). That is, $a^2 + b^2 = c^2$.

We need to find two numbers $a$ and $b$ such that $a^2 + b^2 = 13$. We can look for integer squares that add up to 13.

The squares of integers are $0^2=0$, $1^2=1$, $2^2=4$, $3^2=9$, $4^2=16$, etc.

We can see that $4 + 9 = 13$. So, we can choose $a^2 = 4$ and $b^2 = 9$, which means $a = \sqrt{4} = 2$ and $b = \sqrt{9} = 3$.

Thus, $\sqrt{13}$ is the length of the hypotenuse of a right-angled triangle with legs of length 2 units and 3 units.

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Construction:

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1. Draw a horizontal number line and mark the origin O at point 0.

2. From the origin O, move 2 units to the right along the number line and mark this point as A. So, the length of OA is 2 units.

3. At point A, draw a line segment AB perpendicular to the number line, such that the length of AB is 3 units.

4. Connect point O to point B. We now have a right-angled triangle OAB, with the right angle at A.

5. By the Pythagorean theorem in $\triangle$OAB:

$OB^2 = OA^2 + AB^2$

$OB^2 = 2^2 + 3^2$

$OB^2 = 4 + 9$

$OB^2 = 13$

$OB = \sqrt{13}$

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6. Using a compass, place the needle at the origin O and set the radius equal to the length of OB ($\sqrt{13}$ units).

7. Draw an arc that intersects the positive side of the number line. Mark the point of intersection as P.

8. The distance OP is equal to the radius of the arc, which is OB.

Thus, the distance OP is $\sqrt{13}$ units.

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The point P on the number line represents the location of $\sqrt{13}$.

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Alternate Construction:

We could also take OA = 3 units along the number line and draw a perpendicular AB = 2 units. The hypotenuse OB would still have length $\sqrt{3^2 + 2^2} = \sqrt{9+4} = \sqrt{13}$. The rest of the steps would be the same.

Solution:

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Let the given number be $x$.

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The non-repeating part is '12' (two digits). Multiply equation (i) by $10^2 = 100$ to move the decimal point past the non-repeating digits.

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The repeating part is '3' (one digit). Multiply equation (ii) by $10^1 = 10$ to move the decimal point past one repeating block.

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Subtract equation (ii) from equation (iii) to eliminate the repeating decimal part.

$\begin{array}{rc} & 1000x & = & 123.333... \\ - & 100x & = & \phantom{0}12.333... \\ \hline & 900x & = & 111.000... \\ \hline \end{array}$

So, we have $900x = 111$.

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Solve for $x$ by dividing both sides by 900.

$x = \frac{111}{900}$

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Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both 111 and 900 are divisible by 3.

$\frac{111 \div 3}{900 \div 3} = \frac{37}{300}$

The number 37 is a prime number, and 300 is not divisible by 37. So, the fraction is in its simplest form.

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Thus, $0.12\overline{3}$ expressed in the form $\frac{p}{q}$ is $\frac{37}{300}$, where $p=37$ and $q=300$ are integers and $q \neq 0$.

Solution:

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The given expression is $(3\sqrt{5} - 5\sqrt{2})(4\sqrt{5} + 3\sqrt{2})$.

---

We can simplify this expression by multiplying the terms using the distributive property (often remembered by the acronym FOIL - First, Outer, Inner, Last).

$(3\sqrt{5} - 5\sqrt{2})(4\sqrt{5} + 3\sqrt{2}) = (3\sqrt{5})(4\sqrt{5}) + (3\sqrt{5})(3\sqrt{2}) + (-5\sqrt{2})(4\sqrt{5}) + (-5\sqrt{2})(3\sqrt{2})$

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Let's calculate each term:

First terms: $(3\sqrt{5})(4\sqrt{5}) = (3 \times 4) \times (\sqrt{5} \times \sqrt{5}) = 12 \times 5 = 60$

Outer terms: $(3\sqrt{5})(3\sqrt{2}) = (3 \times 3) \times (\sqrt{5} \times \sqrt{2}) = 9 \times \sqrt{10} = 9\sqrt{10}$

Inner terms: $(-5\sqrt{2})(4\sqrt{5}) = (-5 \times 4) \times (\sqrt{2} \times \sqrt{5}) = -20 \times \sqrt{10} = -20\sqrt{10}$

Last terms: $(-5\sqrt{2})(3\sqrt{2}) = (-5 \times 3) \times (\sqrt{2} \times \sqrt{2}) = -15 \times 2 = -30$

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Now, add these terms together:

$(3\sqrt{5} - 5\sqrt{2})(4\sqrt{5} + 3\sqrt{2}) = 60 + 9\sqrt{10} - 20\sqrt{10} - 30$

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Combine the like terms (rational terms and terms with $\sqrt{10}$):

$= (60 - 30) + (9\sqrt{10} - 20\sqrt{10})$

$= 30 + (9 - 20)\sqrt{10}$

$= 30 - 11\sqrt{10}$

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The simplified expression is $30 - 11\sqrt{10}$.

Solution:

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The given equation is: $$\frac{6}{3\sqrt{2} \;-\; 2\sqrt{3}} = 3\sqrt{2} - a\sqrt{3}$$

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To find the value of $a$, we first simplify the left side of the equation by rationalizing the denominator.

The denominator is $3\sqrt{2} \;-\; 2\sqrt{3}$. The conjugate of the denominator is $3\sqrt{2} \;+\; 2\sqrt{3}$.

---

Multiply the numerator and the denominator of the left side by the conjugate:

$$\frac{6}{3\sqrt{2} \;-\; 2\sqrt{3}} = \frac{6}{(3\sqrt{2} \;-\; 2\sqrt{3})} \times \frac{(3\sqrt{2} \;+\; 2\sqrt{3})}{(3\sqrt{2} \;+\; 2\sqrt{3})}$$

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Simplify the numerator:

$6(3\sqrt{2} \;+\; 2\sqrt{3}) = 6 \times 3\sqrt{2} \;+\; 6 \times 2\sqrt{3} = 18\sqrt{2} \;+\; 12\sqrt{3}$

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Simplify the denominator using the difference of squares formula $(A-B)(A+B) = A^2 - B^2$, where $A = 3\sqrt{2}$ and $B = 2\sqrt{3}$:

$$(3\sqrt{2} \;-\; 2\sqrt{3})(3\sqrt{2} \;+\; 2\sqrt{3}) = (3\sqrt{2})^2 \;-\; (2\sqrt{3})^2$$

Calculate the squares:

$(3\sqrt{2})^2 = 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18$

$(2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$

So, the denominator is $18 \;-\; 12 = 6$.

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The simplified left side of the equation is:

$$\frac{18\sqrt{2} \;+\; 12\sqrt{3}}{6}$$

Divide each term in the numerator by the denominator:

$$\frac{18\sqrt{2}}{6} \;+\; \frac{12\sqrt{3}}{6} = 3\sqrt{2} \;+\; 2\sqrt{3}$$

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Now, set the simplified left side equal to the right side of the original equation:

$$3\sqrt{2} \;+\; 2\sqrt{3} = 3\sqrt{2} \;-\; a\sqrt{3}$$

---

Subtract $3\sqrt{2}$ from both sides:

$2\sqrt{3} = -a\sqrt{3}$

---

Divide both sides by $\sqrt{3}$ (since $\sqrt{3} \neq 0$):

$$2 = -a$$

---

Multiply both sides by -1 to solve for $a$:

$$a = -2$$

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The value of $a$ is $-2$.

Solution:

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The given expression is $\left[ 5\left( 8^{\frac{1}{3}} + 27^{\frac{1}{3}} \right)^3 \right]^{\frac{1}{4}}$.

---

We simplify the expression following the order of operations (PEMDAS/BODMAS).

---

First, simplify the terms inside the innermost parentheses:

$8^{\frac{1}{3}}$ is the cube root of 8.

$8 = 2^3$, so $8^{\frac{1}{3}} = (2^3)^{\frac{1}{3}} = 2^{3 \times \frac{1}{3}} = 2^1 = 2$.

$27^{\frac{1}{3}}$ is the cube root of 27.

$27 = 3^3$, so $27^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = 3^{3 \times \frac{1}{3}} = 3^1 = 3$.

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Substitute these values back into the expression inside the first set of parentheses:

$\left( 8^{\frac{1}{3}} + 27^{\frac{1}{3}} \right) = (2 + 3) = 5$

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Now, the expression inside the square brackets becomes:

$\left[ 5\left( 5 \right)^3 \right]$

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Calculate the term raised to the power of 3:

$5^3 = 5 \times 5 \times 5 = 125$.

---

Substitute this value back into the expression inside the square brackets:

$\left[ 5 \times 125 \right]$

---

Multiply the numbers inside the square brackets:

$5 \times 125 = 625$.

---

The expression is now $\left[ 625 \right]^{\frac{1}{4}}$.

Raising to the power of $\frac{1}{4}$ is the same as taking the fourth root.

$(625)^{\frac{1}{4}} = \sqrt[4]{625}$.

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We need to find a number that, when multiplied by itself four times, equals 625. We can check powers of integers:

$1^4 = 1$

$2^4 = 16$

$3^4 = 81$

$4^4 = 256$

$5^4 = 5 \times 5 \times 5 \times 5 = 25 \times 25 = 625$.

So, $\sqrt[4]{625} = 5$.

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The value of the expression is 5.

Solution:

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We are asked to classify the variables x, y, z, and u as rational or irrational based on the given equations.

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(i) $x^2 = 5$

Solving for x, we get $x = \pm\sqrt{5}$.

The number 5 is not a perfect square (i.e., there is no integer $k$ such that $k^2 = 5$). The square root of a positive integer that is not a perfect square is an irrational number.

Therefore, $\sqrt{5}$ is an irrational number, and $-\sqrt{5}$ is also an irrational number.

The variable x represents an irrational number.

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(ii) $y^2 = 9$

Solving for y, we get $y = \pm\sqrt{9}$.

Since $3^2 = 9$, $\sqrt{9} = 3$.

So, $y = \pm 3$.

The numbers 3 and -3 are integers. Integers can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $3 = \frac{3}{1}$, $-3 = \frac{-3}{1}$).

Therefore, 3 and -3 are rational numbers.

The variable y represents a rational number.

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(iii) $z^2 = 0.04$

Solving for z, we get $z = \pm\sqrt{0.04}$.

We can write 0.04 as a fraction: $0.04 = \frac{4}{100}$.

So, $z = \pm\sqrt{\frac{4}{100}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$z = \pm \frac{\sqrt{4}}{\sqrt{100}} = \pm \frac{2}{10}$

Simplifying the fraction: $\frac{2}{10} = \frac{1}{5}$.

So, $z = \pm \frac{1}{5}$ (or $\pm 0.2$).

The numbers $\frac{1}{5}$ and $-\frac{1}{5}$ are in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g., $p=1, q=5$ or $p=-1, q=5$).

Therefore, $\frac{1}{5}$ and $-\frac{1}{5}$ are rational numbers.

The variable z represents a rational number.

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(iv) $u^2 = \frac{17}{4}$

Solving for u, we get $u = \pm\sqrt{\frac{17}{4}}$.

Using the property $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$:

$u = \pm \frac{\sqrt{17}}{\sqrt{4}} = \pm \frac{\sqrt{17}}{2}$.

The number 17 is not a perfect square, so $\sqrt{17}$ is an irrational number.

The number $\frac{\sqrt{17}}{2}$ is the quotient of an irrational number ($\sqrt{17}$) and a non-zero rational number (2). The quotient of an irrational number and a non-zero rational number is always irrational.

Therefore, $\frac{\sqrt{17}}{2}$ is an irrational number, and $-\frac{\sqrt{17}}{2}$ is also an irrational number.

The variable u represents an irrational number.

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In summary:

x is irrational.

y is rational.

z is rational.

u is irrational.

Solution:

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To find rational numbers between two given rational numbers, we can use various methods, such as finding the average of the two numbers repeatedly or by expressing the numbers with a common denominator and then finding fractions between them.

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(i) Between –1 and –2

The given numbers are -1 and -2. Both are rational numbers.

Let's find three rational numbers between them.

Method 1: Using Decimal Form

-1 can be written as -1.0.

-2 can be written as -2.0.

We need numbers between -2.0 and -1.0. We can easily list terminating decimals between them, which are rational numbers. For example:

-1.1, -1.2, -1.3, -1.4, -1.5, -1.6, -1.7, -1.8, -1.9.

Let's choose three of these:

$-1.1 = -\frac{11}{10}$

$-1.5 = -\frac{15}{10} = -\frac{3}{2}$

$-1.9 = -\frac{19}{10}$

Three rational numbers between –1 and –2 are $-\frac{11}{10}$, $-\frac{3}{2}$, $-\frac{19}{10}$.

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(ii) Between 0.1 and 0.11

The given numbers are 0.1 and 0.11. Both are terminating decimals, so they are rational numbers.

0.1 can be written as 0.100.

0.11 can be written as 0.110.

We need numbers between 0.100 and 0.110. We can list terminating decimals between them:

0.101, 0.102, 0.103, 0.104, ..., 0.109.

Let's choose three of these:

$0.101 = \frac{101}{1000}$

$0.105 = \frac{105}{1000} = \frac{21}{200}$

$0.109 = \frac{109}{1000}$

Three rational numbers between 0.1 and 0.11 are $\frac{101}{1000}$, $\frac{21}{200}$, $\frac{109}{1000}$.

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(iii) Between $\frac{5}{7}$ and $\frac{6}{7}$

The given numbers are $\frac{5}{7}$ and $\frac{6}{7}$. Both are rational numbers.

Method: Increase the denominator.

To find rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$, we can multiply the numerator and denominator of both fractions by a number greater than 1, say 10.

$\frac{5}{7} = \frac{5 \times 10}{7 \times 10} = \frac{50}{70}$

$\frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70}$

Now we can easily find fractions with denominator 70 and numerators between 50 and 60.

For example, $\frac{51}{70}, \frac{52}{70}, \frac{53}{70}, \dots, \frac{59}{70}$.

Let's choose three of these:

$\frac{51}{70}$

$\frac{55}{70} = \frac{11}{14}$ (simplified)

$\frac{59}{70}$

Three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70}$, $\frac{11}{14}$, $\frac{59}{70}$.

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(iv) Between $\frac{1}{4}$ and $\frac{1}{5}$

The given numbers are $\frac{1}{4}$ and $\frac{1}{5}$. Both are rational numbers.

Method: Find a common denominator and then increase it.

The least common multiple of the denominators 4 and 5 is 20. Express both fractions with denominator 20:

$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$

$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$

We need numbers between $\frac{4}{20}$ and $\frac{5}{20}$. Multiply numerator and denominator of both fractions by a number greater than 1, say 10, to create more space between the numerators.

$\frac{4}{20} = \frac{4 \times 10}{20 \times 10} = \frac{40}{200}$

$\frac{5}{20} = \frac{5 \times 10}{20 \times 10} = \frac{50}{200}$

Now we can find fractions with denominator 200 and numerators between 40 and 50.

For example, $\frac{41}{200}, \frac{42}{200}, \frac{43}{200}, \dots, \frac{49}{200}$.

Let's choose three of these:

$\frac{41}{200}$

$\frac{45}{200} = \frac{9}{40}$ (simplified)

$\frac{49}{200}$

Three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{41}{200}$, $\frac{9}{40}$, $\frac{49}{200}$.

Solution:

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We need to insert one rational number and one irrational number between each given pair of numbers.

Recall that a rational number can be expressed as a terminating or non-terminating repeating decimal. An irrational number has a non-terminating and non-repeating decimal expansion.

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(i) 2 and 3

Rational number: We can take the average of 2 and 3: $\frac{2+3}{2} = \frac{5}{2} = 2.5$. $2 < 2.5 < 3$. Since 2.5 is a terminating decimal, it is a rational number.

Irrational number: We can construct a non-terminating, non-repeating decimal between 2 and 3. For example, $2.1010010001...$ (where the number of zeros between consecutive '1's increases by one each time). This number is greater than 2 and less than 3. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

---

(ii) 0 and 0.1

Rational number: We can choose a terminating decimal between 0 and 0.1, for example, 0.05. $0 < 0.05 < 0.1$. Since 0.05 is a terminating decimal, it is a rational number ($0.05 = \frac{5}{100} = \frac{1}{20}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0 and 0.1. For example, $0.01010010001...$ (where the number of zeros between consecutive '1's increases by one each time). This number is greater than 0 and less than 0.1. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

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(iii) $\frac{1}{3}$ and $\frac{1}{2}$

We can approximate the decimal values: $\frac{1}{3} \approx 0.333...$ and $\frac{1}{2} = 0.5$. We need a number between 0.333... and 0.5.

Rational number: We can choose a terminating decimal like 0.4. $0.333... < 0.4 < 0.5$. Since 0.4 is a terminating decimal, it is rational ($0.4 = \frac{4}{10} = \frac{2}{5}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.333... and 0.5. For example, $0.414114111...$ (where the number of '1's between consecutive '4's increases by one each time). This number is greater than 0.333... and less than 0.5. Since its decimal expansion is non-terminating and non-repeating, it is an irrational number.

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(iv) $\frac{-2}{5}$ and $\frac{1}{2}$

We can convert these to decimals: $\frac{-2}{5} = -0.4$ and $\frac{1}{2} = 0.5$. We need a number between -0.4 and 0.5.

Rational number: We can choose an integer like 0. $-0.4 < 0 < 0.5$. Since 0 is an integer, it is rational ($0 = \frac{0}{1}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between -0.4 and 0.5. For example, $0.1010010001...$ (pattern of increasing zeros between 1s). $-0.4 < 0.101001... < 0.5$. This decimal is non-terminating and non-repeating, so it is irrational.

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(v) 0.15 and 0.16

We need a number between 0.15 and 0.16.

Rational number: We can choose a terminating decimal like 0.153. $0.15 < 0.153 < 0.16$. Since 0.153 is a terminating decimal, it is rational ($0.153 = \frac{153}{1000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.15 and 0.16. For example, $0.15110111001111000...$ (pattern of increasing 1s followed by increasing 0s). $0.15 < 0.1511011100... < 0.16$. This decimal is non-terminating and non-repeating, so it is irrational.

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(vi) $\sqrt{2}$ and $\sqrt{3}$

We can approximate the decimal values: $\sqrt{2} \approx 1.414...$ and $\sqrt{3} \approx 1.732...$. We need a number between 1.414... and 1.732....

Rational number: We can choose a terminating decimal like 1.5. $\sqrt{2} \approx 1.414 < 1.5 < 1.732 \approx \sqrt{3}$. Since 1.5 is a terminating decimal, it is rational ($1.5 = \frac{15}{10} = \frac{3}{2}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 1.414... and 1.732.... For example, $1.5050050005...$ (pattern of increasing zeros between 5s). $\sqrt{2} < 1.505005... < \sqrt{3}$. This decimal is non-terminating and non-repeating, so it is irrational.

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(vii) 2.357 and 3.121

These are terminating decimals (and thus rational numbers). We need a number between 2.357 and 3.121.

Rational number: We can choose a terminating decimal like 2.5. $2.357 < 2.5 < 3.121$. Since 2.5 is a terminating decimal, it is rational ($2.5 = \frac{25}{10} = \frac{5}{2}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 2.357 and 3.121. For example, $2.414114111...$ (pattern of increasing 1s between 4s). $2.357 < 2.414114... < 3.121$. This decimal is non-terminating and non-repeating, so it is irrational.

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(viii) 0.0001 and 0.001

These are terminating decimals (and thus rational numbers). We need a number between 0.0001 and 0.001.

Rational number: We can choose a terminating decimal like 0.0005. $0.0001 < 0.0005 < 0.001$. Since 0.0005 is a terminating decimal, it is rational ($0.0005 = \frac{5}{10000} = \frac{1}{2000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.0001 and 0.001. For example, $0.000120120012000...$ (pattern of '12' followed by increasing zeros). $0.0001 < 0.0001201200... < 0.001$. This decimal is non-terminating and non-repeating, so it is irrational.

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(ix) 3.623623 and 0.484848

The numbers are 3.623623 and 0.484848. Both are terminating decimals (and thus rational numbers). Note that $0.484848 < 3.623623$. We need a number between them.

Rational number: We can choose a terminating decimal like 1. $0.484848 < 1 < 3.623623$. Since 1 is an integer, it is rational ($1 = \frac{1}{1}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 0.484848 and 3.623623. For example, $1.121121112...$ (pattern of increasing 1s between 2s). $0.484848 < 1.121121112... < 3.623623$. This decimal is non-terminating and non-repeating, so it is irrational.

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(x) 6.375289 and 6.375738

These are terminating decimals (and thus rational numbers). We need a number between 6.375289 and 6.375738.

Rational number: We can choose a terminating decimal like 6.3753. $6.375289 < 6.3753 < 6.375738$. Since 6.3753 is a terminating decimal, it is rational ($6.3753 = \frac{63753}{10000}$).

Irrational number: We can construct a non-terminating, non-repeating decimal between 6.375289 and 6.375738. For example, $6.3754040040004...$ (pattern of increasing zeros between 4s). $6.375289 < 6.375404004... < 6.375738$. This decimal is non-terminating and non-repeating, so it is irrational.

Solution:

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To represent the given numbers on the number line, we first need to understand their values and positions relative to zero and other integers.

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The given numbers are: 7, 7.2, $\frac{-3}{2}$, and $\frac{-12}{5}$.

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Let's determine the approximate value or form of each number:

• 7 is a positive integer.
• 7.2 is a positive decimal number between 7 and 8.
• $\frac{-3}{2}$ is a negative fraction. Converting it to a decimal: $\frac{-3}{2} = -1.5$. This number is halfway between -1 and -2.
• $\frac{-12}{5}$ is a negative fraction. Converting it to a decimal: $\frac{-12}{5} = -2.4$. This number is between -2 and -3.

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Now, we can draw a number line and mark these points:

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1. Draw a horizontal line. Mark the origin (0) and set a suitable scale (e.g., mark integers 1 unit apart).

2. Mark the integer points clearly, including negative integers.

3. Locate the point for 7 on the positive side of the number line.

4. Locate the point for 7.2. This is slightly to the right of 7, specifically two-tenths of the way between 7 and 8.

5. Locate the point for $\frac{-3}{2}$ (-1.5). This is on the negative side, exactly halfway between -1 and -2.

6. Locate the point for $\frac{-12}{5}$ (-2.4). This is on the negative side, between -2 and -3. It is four-tenths of the way from -2 towards -3.

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A representation on the number line would look like this (conceptually, you would draw this):

---o----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|--->

... -3 -2.4 -2 -1.5 -1 0 1 2 3 4 5 6 7 7.2 8 ...

Mark specific points for $\frac{-12}{5}$, $\frac{-3}{2}$, 7, and 7.2 on your drawn number line with labels.

Solution:

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To locate $\sqrt{5}$, $\sqrt{10}$, and $\sqrt{17}$ on the number line, we use the Pythagorean theorem ($a^2 + b^2 = c^2$). We find two numbers $a$ and $b$ whose squares add up to the number under the square root. The length of the hypotenuse $c$ will be the desired square root.

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Locating $\sqrt{5}$ on the number line:

We need $a^2 + b^2 = 5$. We can choose $a=1$ and $b=2$, since $1^2 + 2^2 = 1 + 4 = 5$.

Construction steps:

1. Draw a number line and mark the origin O (representing 0).

2. Mark a point A on the number line at a distance of 2 units from O in the positive direction. OA = 2 units.

3. At point A, construct a line segment AB perpendicular to OA, with length 1 unit.

4. Join OB. $\triangle$OAB is a right-angled triangle with the right angle at A.

5. By the Pythagorean theorem, $OB^2 = OA^2 + AB^2 = 2^2 + 1^2 = 4 + 1 = 5$. So, $OB = \sqrt{5}$.

6. With O as the centre and radius OB, draw an arc intersecting the number line at point P. The point P represents $\sqrt{5}$.

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Locating $\sqrt{10}$ on the number line:

We need $a^2 + b^2 = 10$. We can choose $a=1$ and $b=3$, since $1^2 + 3^2 = 1 + 9 = 10.

Construction steps:

1. Draw a number line and mark the origin O (representing 0).

2. Mark a point C on the number line at a distance of 3 units from O in the positive direction. OC = 3 units.

3. At point C, construct a line segment CD perpendicular to OC, with length 1 unit.

4. Join OD. $\triangle$OCD is a right-angled triangle with the right angle at C.

5. By the Pythagorean theorem, $OD^2 = OC^2 + CD^2 = 3^2 + 1^2 = 9 + 1 = 10$. So, $OD = \sqrt{10}$.

6. With O as the centre and radius OD, draw an arc intersecting the number line at point Q. The point Q represents $\sqrt{10}$.

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Locating $\sqrt{17}$ on the number line:

We need $a^2 + b^2 = 17$. We can choose $a=1$ and $b=4$, since $1^2 + 4^2 = 1 + 16 = 17.

Construction steps:

1. Draw a number line and mark the origin O (representing 0).

2. Mark a point E on the number line at a distance of 4 units from O in the positive direction. OE = 4 units.

3. At point E, construct a line segment EF perpendicular to OE, with length 1 unit.

4. Join OF. $\triangle$OEF is a right-angled triangle with the right angle at E.

5. By the Pythagorean theorem, $OF^2 = OE^2 + EF^2 = 4^2 + 1^2 = 16 + 1 = 17$. So, $OF = \sqrt{17}$.

6. With O as the centre and radius OF, draw an arc intersecting the number line at point R. The point R represents $\sqrt{17}$.

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(i) Representing $\sqrt{4.5}$ on the number line:

Construction Steps:

1. Draw a line and mark a point A on it. Let A represent the number 0.

2. Mark a point B on the line such that the distance AB is 4.5 units. The point B corresponds to the number 4.5 on the number line relative to A being 0.

3. Extend the line segment AB to the right to a point C such that the distance BC is 1 unit. The total distance AC is $4.5 + 1 = 5.5$ units.

4. Find the midpoint of the line segment AC. Let this midpoint be O. The distance $OC = OA = \frac{AC}{2} = \frac{5.5}{2} = 2.75$ units.

5. With O as the center and radius OC (or OA), draw a semicircle.

6. Draw a line perpendicular to the line AC passing through point B. Let this perpendicular line intersect the semicircle at point D.

7. The length of the line segment BD is equal to $\sqrt{4.5}$. (This can be proven using the geometric mean theorem in the right triangle $\triangle ADC$ or by coordinates: If A=0, C=5.5, B=4.5, O=2.75, D is on the semicircle, $OD^2 = OB^2 + BD^2$. $OD = 2.75$, $OB = |4.5 - 2.75| = 1.75$. $BD^2 = OD^2 - OB^2 = (2.75)^2 - (1.75)^2 = (2.75 - 1.75)(2.75 + 1.75) = (1)(4.5) = 4.5$. So $BD = \sqrt{4.5}$.)

8. To represent the length $\sqrt{4.5}$ on the number line starting from the origin A (0), take A as the center and BD as the radius. Draw an arc that intersects the number line at a point E.

9. The point E on the number line represents the value $\sqrt{4.5}$.

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(ii) Representing $\sqrt{5.6}$ on the number line:

Construction Steps:

1. Draw a line and mark a point A on it. Let A represent the number 0.

2. Mark a point B on the line such that the distance AB is 5.6 units. The point B corresponds to the number 5.6.

3. Extend the line segment AB to a point C such that the distance BC is 1 unit. The total distance AC is $5.6 + 1 = 6.6$ units.

4. Find the midpoint of the line segment AC. Let this midpoint be O. The distance $OC = \frac{AC}{2} = \frac{6.6}{2} = 3.3$ units.

5. With O as the center and radius OC, draw a semicircle.

6. Draw a line perpendicular to the line AC passing through point B. Let this perpendicular line intersect the semicircle at point D.

7. The length of the line segment BD is equal to $\sqrt{5.6}$.

8. To represent the length $\sqrt{5.6}$ on the number line starting from A (0), take A as the center and BD as the radius. Draw an arc that intersects the number line at a point E.

9. The point E on the number line represents the value $\sqrt{5.6}$.

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(iii) Representing $\sqrt{8.1}$ on the number line:

Construction Steps:

1. Draw a line and mark a point A on it. Let A represent the number 0.

2. Mark a point B on the line such that the distance AB is 8.1 units. The point B corresponds to the number 8.1.

3. Extend the line segment AB to a point C such that the distance BC is 1 unit. The total distance AC is $8.1 + 1 = 9.1$ units.

4. Find the midpoint of the line segment AC. Let this midpoint be O. The distance $OC = \frac{AC}{2} = \frac{9.1}{2} = 4.55$ units.

5. With O as the center and radius OC, draw a semicircle.

6. Draw a line perpendicular to the line AC passing through point B. Let this perpendicular line intersect the semicircle at point D.

7. The length of the line segment BD is equal to $\sqrt{8.1}$.

8. To represent the length $\sqrt{8.1}$ on the number line starting from A (0), take A as the center and BD as the radius. Draw an arc that intersects the number line at a point E.

9. The point E on the number line represents the value $\sqrt{8.1}$.

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(iv) Representing $\sqrt{2.3}$ on the number line:

Construction Steps:

1. Draw a line and mark a point A on it. Let A represent the number 0.

2. Mark a point B on the line such that the distance AB is 2.3 units. The point B corresponds to the number 2.3.

3. Extend the line segment AB to a point C such that the distance BC is 1 unit. The total distance AC is $2.3 + 1 = 3.3$ units.

4. Find the midpoint of the line segment AC. Let this midpoint be O. The distance $OC = \frac{AC}{2} = \frac{3.3}{2} = 1.65$ units.

5. With O as the center and radius OC, draw a semicircle.

6. Draw a line perpendicular to the line AC passing through point B. Let this perpendicular line intersect the semicircle at point D.

7. The length of the line segment BD is equal to $\sqrt{2.3}$.

8. To represent the length $\sqrt{2.3}$ on the number line starting from A (0), take A as the center and BD as the radius. Draw an arc that intersects the number line at a point E.

9. The point E on the number line represents the value $\sqrt{2.3}$.

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(i) Express 0.2 in the form $\frac{p}{q}$:

$0.2 = \frac{2}{10}$

Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2.

$0.2 = \frac{\cancel{2}^1}{\cancel{10}_5}$

Thus, $0.2 = \frac{1}{5}$.

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(ii) Express 0.888... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since one digit is repeating after the decimal point, multiply equation (i) by 10.

Subtract equation (i) from equation (ii).

Divide both sides by 9.

Thus, $0.888... = \frac{8}{9}$.

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(iii) Express 5.$\overline{2}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since one digit is repeating after the decimal point, multiply equation (i) by 10.

Subtract equation (i) from equation (ii).

Divide both sides by 9.

Thus, 5.$\overline{2}$ = $\frac{47}{9}$.

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(iv) Express 0.$\overline{001}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since three digits are repeating after the decimal point, multiply equation (i) by $10^3 = 1000$.

Subtract equation (i) from equation (ii).

Divide both sides by 999.

Thus, 0.$\overline{001}$ = $\frac{1}{999}$.

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(v) Express 0.2555... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there is one non-repeating digit (2) after the decimal point, multiply equation (i) by 10.

Since there is one repeating digit (5) after the decimal point in equation (ii), multiply equation (ii) by 10.

Subtract equation (ii) from equation (iii).

Divide both sides by 90.

Thus, 0.2555... = $\frac{23}{90}$.

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(vi) Express 0.1$\overline{34}$ in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there is one non-repeating digit (1) after the decimal point, multiply equation (i) by 10.

Since there are two repeating digits (34) after the decimal point in equation (ii), multiply equation (ii) by $10^2 = 100$.

Subtract equation (ii) from equation (iii).

Divide both sides by 990.

Thus, 0.1$\overline{34}$ = $\frac{133}{990}$.

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(vii) Express 0.00323232… in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since there are two non-repeating digits (00) after the decimal point, multiply equation (i) by $10^2 = 100$.

Since there are two repeating digits (32) after the decimal point in equation (ii), multiply equation (ii) by $10^2 = 100$.

Subtract equation (ii) from equation (iii).

Divide both sides by 9900.

Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4.

Thus, 0.00323232… = $\frac{8}{2475}$.

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(viii) Express 0.404040... in the form $\frac{p}{q}$:

Let $x$ be the given number.

Since two digits are repeating after the decimal point, multiply equation (i) by $10^2 = 100$.

Subtract equation (i) from equation (ii).

Divide both sides by 99.

Thus, 0.404040... = $\frac{40}{99}$.

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Let $x$ be the given number.

We can write this as:

The repeating block of digits is $142857$, which consists of 6 digits.

Multiply equation (i) by $10^6 = 1000000$.

We can write this as:

Subtract equation (i) from equation (ii).

Now, solve for $x$ by dividing both sides by 999999.

To simplify the fraction, we can observe that $142857 \times 7 = 999999$.

So, we can divide both the numerator and the denominator by 142857.

Thus, we have shown that $0.142857142857... = \frac{1}{7}$.

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(i) Simplify $\sqrt{45}$ – 3$\sqrt{20}$ + 4$\sqrt{5}$:

First, simplify the individual square roots:

$\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5}$

$\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$

Substitute these simplified radicals back into the expression:

$3\sqrt{5} - 3(2\sqrt{5}) + 4\sqrt{5}$

$= 3\sqrt{5} - 6\sqrt{5} + 4\sqrt{5}$

Combine the terms as they all have the same radical $\sqrt{5}$:

$= (3 - 6 + 4)\sqrt{5}$

$= (1)\sqrt{5}$

$= \sqrt{5}$

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(ii) Simplify $\frac{\sqrt{24}}{8}$ + $\frac{\sqrt{54}}{9}$:

First, simplify the square roots in the numerators:

$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$

$\sqrt{54} = \sqrt{9 \times 6} = \sqrt{9} \times \sqrt{6} = 3\sqrt{6}$

Substitute these into the expression:

$\frac{2\sqrt{6}}{8} + \frac{3\sqrt{6}}{9}$

Simplify the fractions:

$\frac{\cancel{2}^1\sqrt{6}}{\cancel{8}_4} + \frac{\cancel{3}^1\sqrt{6}}{\cancel{9}_3}$

$= \frac{\sqrt{6}}{4} + \frac{\sqrt{6}}{3}$

To add these fractions, find a common denominator, which is the LCM of 4 and 3, i.e., 12.

$= \frac{3 \times \sqrt{6}}{3 \times 4} + \frac{4 \times \sqrt{6}}{4 \times 3}$

$= \frac{3\sqrt{6}}{12} + \frac{4\sqrt{6}}{12}$

Combine the terms:

$= \frac{3\sqrt{6} + 4\sqrt{6}}{12}$

$= \frac{(3+4)\sqrt{6}}{12}$

$= \frac{7\sqrt{6}}{12}$

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(iii) Simplify $\sqrt [4] {12} \times \sqrt [7] {6}$:

To multiply radicals with different indices, we need to express them with a common index. The indices are 4 and 7. The LCM of 4 and 7 is 28.

Convert $\sqrt[4]{12}$ to a 28th root:

$\sqrt[4]{12} = 12^{1/4} = 12^{7/28} = \sqrt[28]{12^7}$

Convert $\sqrt[7]{6}$ to a 28th root:

$\sqrt[7]{6} = 6^{1/7} = 6^{4/28} = \sqrt[28]{6^4}$

Now multiply the 28th roots:

$\sqrt[28]{12^7} \times \sqrt[28]{6^4} = \sqrt[28]{12^7 \times 6^4}$

Express the bases 12 and 6 in terms of their prime factors:

$12 = 2^2 \times 3$

$6 = 2 \times 3$

Substitute these into the expression under the radical:

$12^7 \times 6^4 = (2^2 \times 3)^7 \times (2 \times 3)^4$

$= (2^{2 \times 7} \times 3^7) \times (2^4 \times 3^4)$

$= (2^{14} \times 3^7) \times (2^4 \times 3^4)$

Combine the terms with the same base by adding their exponents:

$= 2^{14+4} \times 3^{7+4}$

$= 2^{18} \times 3^{11}$

So, the expression becomes:

$\sqrt[28]{2^{18} \times 3^{11}}$

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(iv) Simplify $4\sqrt{28}$ ÷ $3\sqrt{7}$ ÷ $\sqrt [3] {7}$:

First, simplify the square root in the first term:

$\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}$

Substitute this back into the expression:

$4(2\sqrt{7})$ ÷ $3\sqrt{7}$ ÷ $\sqrt [3] {7}$

$= 8\sqrt{7}$ ÷ $3\sqrt{7}$ ÷ $\sqrt [3] {7}$

Division can be written as fractions:

$= \frac{8\sqrt{7}}{3\sqrt{7}} \div \sqrt[3]{7}$

Simplify the first fraction:

$= \frac{8\cancel{\sqrt{7}}}{3\cancel{\sqrt{7}}} = \frac{8}{3}$

Now the expression is:

$= \frac{8}{3} \div \sqrt[3]{7}$

$= \frac{8}{3 \times \sqrt[3]{7}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt[3]{7^2} = \sqrt[3]{49}$:

$= \frac{8}{3\sqrt[3]{7}} \times \frac{\sqrt[3]{49}}{\sqrt[3]{49}}$

$= \frac{8\sqrt[3]{49}}{3 \times \sqrt[3]{7 \times 49}}$

$= \frac{8\sqrt[3]{49}}{3 \times \sqrt[3]{343}}$

Since $\sqrt[3]{343} = 7$ (because $7^3 = 343$):

$= \frac{8\sqrt[3]{49}}{3 \times 7}$

$= \frac{8\sqrt[3]{49}}{21}$

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(v) Simplify $3\sqrt{3}$ + $2\sqrt{27}$ + $\frac{7}{\sqrt{3}}$:

First, simplify $\sqrt{27}$ and rationalize the third term:

$\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$

Rationalize $\frac{7}{\sqrt{3}}$ by multiplying the numerator and denominator by $\sqrt{3}$:

$\frac{7}{\sqrt{3}} = \frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{7\sqrt{3}}{3}$

Substitute these back into the expression:

$3\sqrt{3} + 2(3\sqrt{3}) + \frac{7\sqrt{3}}{3}$

$= 3\sqrt{3} + 6\sqrt{3} + \frac{7\sqrt{3}}{3}$

Combine the terms with $\sqrt{3}$. Find a common denominator, which is 3.

$= \frac{3 \times 3\sqrt{3}}{3} + \frac{3 \times 6\sqrt{3}}{3} + \frac{7\sqrt{3}}{3}$

$= \frac{9\sqrt{3}}{3} + \frac{18\sqrt{3}}{3} + \frac{7\sqrt{3}}{3}$

Add the numerators:

$= \frac{9\sqrt{3} + 18\sqrt{3} + 7\sqrt{3}}{3}$

$= \frac{(9+18+7)\sqrt{3}}{3}$

$= \frac{34\sqrt{3}}{3}$

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(vi) Simplify $\left( \sqrt{3} - \sqrt{2} \right)^{2}$:

Use the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$, where $a = \sqrt{3}$ and $b = \sqrt{2}$.

$(\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2$

$= 3 - 2\sqrt{3 \times 2} + 2$

$= 3 - 2\sqrt{6} + 2$

Combine the constant terms:

$= (3 + 2) - 2\sqrt{6}$

$= 5 - 2\sqrt{6}$

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(vii) Simplify $\sqrt [4] {81}$ - $8\sqrt [3] {216}$ + $15\sqrt [5] {32}$ + $\sqrt{225}$:

Evaluate each radical term:

$\sqrt[4]{81}$: Find a number that, when multiplied by itself 4 times, equals 81. $3^4 = 3 \times 3 \times 3 \times 3 = 81$. So, $\sqrt[4]{81} = 3$.

$\sqrt[3]{216}$: Find a number that, when multiplied by itself 3 times, equals 216. $6^3 = 6 \times 6 \times 6 = 216$. So, $\sqrt[3]{216} = 6$.

$\sqrt[5]{32}$: Find a number that, when multiplied by itself 5 times, equals 32. $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$. So, $\sqrt[5]{32} = 2$.

$\sqrt{225}$: Find a number that, when multiplied by itself, equals 225. $15^2 = 15 \times 15 = 225$. So, $\sqrt{225} = 15$.

Substitute these values back into the expression:

$3 - 8(6) + 15(2) + 15$

Perform the multiplications:

$= 3 - 48 + 30 + 15$

Combine the terms:

$= (3 + 30 + 15) - 48$

$= 48 - 48$

$= 0$

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(viii) Simplify $\frac{3}{\sqrt{8}}$ + $\frac{1}{\sqrt{2}}$:

Simplify $\sqrt{8}$ in the first term:

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$

Substitute this into the first term:

$\frac{3}{2\sqrt{2}} + \frac{1}{\sqrt{2}}$

Find a common denominator, which is $2\sqrt{2}$. Multiply the second term by $\frac{2}{2}$.

$= \frac{3}{2\sqrt{2}} + \frac{1 \times 2}{\sqrt{2} \times 2}$

$= \frac{3}{2\sqrt{2}} + \frac{2}{2\sqrt{2}}$

Add the fractions:

$= \frac{3 + 2}{2\sqrt{2}}$

$= \frac{5}{2\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$= \frac{5}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$= \frac{5\sqrt{2}}{2 \times (\sqrt{2})^2}$

$= \frac{5\sqrt{2}}{2 \times 2}$

$= \frac{5\sqrt{2}}{4}$

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(ix) Simplify $\frac{2\sqrt{3}}{3} - \frac{\sqrt{3}}{6}$:

The terms already have the same radical, $\sqrt{3}$.

To subtract these fractions, find a common denominator, which is the LCM of 3 and 6, i.e., 6.

Convert the first fraction to have a denominator of 6 by multiplying the numerator and denominator by 2:

$\frac{2\sqrt{3}}{3} = \frac{2\sqrt{3} \times 2}{3 \times 2} = \frac{4\sqrt{3}}{6}$

Now the expression is:

$= \frac{4\sqrt{3}}{6} - \frac{\sqrt{3}}{6}$

Subtract the numerators:

$= \frac{4\sqrt{3} - \sqrt{3}}{6}$

$= \frac{(4-1)\sqrt{3}}{6}$

$= \frac{3\sqrt{3}}{6}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 3.

$= \frac{\cancel{3}^1\sqrt{3}}{\cancel{6}_2}$

$= \frac{\sqrt{3}}{2}$

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(i) Rationalise the denominator of $\frac{2}{3\sqrt{3}}$:

Multiply the numerator and the denominator by $\sqrt{3}$.

$\frac{2}{3\sqrt{3}} = \frac{2}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$= \frac{2\sqrt{3}}{3 \times (\sqrt{3})^2}$

$= \frac{2\sqrt{3}}{3 \times 3}$

$= \frac{2\sqrt{3}}{9}$

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(ii) Rationalise the denominator of $\frac{\sqrt{40}}{\sqrt{3}}$:

Simplify the numerator first: $\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}$.

The expression is $\frac{2\sqrt{10}}{\sqrt{3}}$.

Multiply the numerator and the denominator by $\sqrt{3}$.

$\frac{2\sqrt{10}}{\sqrt{3}} = \frac{2\sqrt{10}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$= \frac{2\sqrt{10 \times 3}}{(\sqrt{3})^2}$

$= \frac{2\sqrt{30}}{3}$

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(iii) Rationalise the denominator of $\frac{3 \;+\; \sqrt{2}}{4\sqrt{2}}$:

Multiply the numerator and the denominator by $\sqrt{2}$.

$\frac{3 + \sqrt{2}}{4\sqrt{2}} = \frac{3 + \sqrt{2}}{4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$= \frac{(3 + \sqrt{2})\sqrt{2}}{4 \times (\sqrt{2})^2}$

Distribute $\sqrt{2}$ in the numerator and simplify the denominator.

$= \frac{3\sqrt{2} + (\sqrt{2})^2}{4 \times 2}$

$= \frac{3\sqrt{2} + 2}{8}$

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(iv) Rationalise the denominator of $\frac{16}{\sqrt{41} \;-\; 5}$:

The denominator is a binomial involving a square root ($\sqrt{41} - 5$). Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{41} + 5$.

$\frac{16}{\sqrt{41} - 5} = \frac{16}{\sqrt{41} - 5} \times \frac{\sqrt{41} + 5}{\sqrt{41} + 5}$

Use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$ in the denominator.

$= \frac{16(\sqrt{41} + 5)}{(\sqrt{41})^2 - 5^2}$

$= \frac{16(\sqrt{41} + 5)}{41 - 25}$

$= \frac{16(\sqrt{41} + 5)}{16}$

Cancel out the common factor 16.

$= \sqrt{41} + 5$

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(v) Rationalise the denominator of $\frac{2 \;+\; \sqrt{3}}{2 \;-\; \sqrt{3}}$:

The denominator is $2 - \sqrt{3}$. Multiply the numerator and the denominator by the conjugate, which is $2 + \sqrt{3}$.

$\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$

Use $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + 2ab + b^2$ in the numerator.

$= \frac{(2 + \sqrt{3})^2}{2^2 - (\sqrt{3})^2}$

$= \frac{2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2}{4 - 3}$

$= \frac{4 + 4\sqrt{3} + 3}{1}$

$= 7 + 4\sqrt{3}$

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(vi) Rationalise the denominator of $\frac{\sqrt{6}}{\sqrt{2} \;+\; \sqrt{3}}$:

The denominator is $\sqrt{2} + \sqrt{3}$. Multiply the numerator and the denominator by the conjugate, which is $\sqrt{2} - \sqrt{3}$.

$\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}$

Use $(a+b)(a-b) = a^2 - b^2$ in the denominator.

$= \frac{\sqrt{6}(\sqrt{2} - \sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2}$

Distribute $\sqrt{6}$ in the numerator.

$= \frac{\sqrt{6 \times 2} - \sqrt{6 \times 3}}{2 - 3}$

$= \frac{\sqrt{12} - \sqrt{18}}{-1}$

Simplify the radicals $\sqrt{12}$ and $\sqrt{18}$.

$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

Substitute these back into the expression.

$= \frac{2\sqrt{3} - 3\sqrt{2}}{-1}$

Divide by -1 (change the sign of each term in the numerator).

$= -(2\sqrt{3} - 3\sqrt{2})$

$= -2\sqrt{3} + 3\sqrt{2}$

Rearrange the terms (optional):

$= 3\sqrt{2} - 2\sqrt{3}$

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(vii) Rationalise the denominator of $\frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$:

The denominator is $\sqrt{3} - \sqrt{2}$. Multiply the numerator and the denominator by the conjugate, which is $\sqrt{3} + \sqrt{2}$.

$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

Use $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + 2ab + b^2$ in the numerator.

$= \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$= \frac{(\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2}{3 - 2}$

$= \frac{3 + 2\sqrt{3 \times 2} + 2}{1}$

$= 3 + 2\sqrt{6} + 2$

$= 5 + 2\sqrt{6}$

---

(viii) Rationalise the denominator of $\frac{3\sqrt{5} \;+\; \sqrt{3}}{\sqrt{5} \;-\; \sqrt{2}}$:

The denominator is $\sqrt{5} - \sqrt{2}$. Multiply the numerator and the denominator by the conjugate, which is $\sqrt{5} + \sqrt{2}$.

$\frac{3\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{2}} = \frac{3\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}$

Use $(a-b)(a+b) = a^2 - b^2$ in the denominator.

$= \frac{(3\sqrt{5} + \sqrt{3})(\sqrt{5} + \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2}$

Multiply the terms in the numerator (using FOIL or distribution).

Numerator: $(3\sqrt{5})(\sqrt{5}) + (3\sqrt{5})(\sqrt{2}) + (\sqrt{3})(\sqrt{5}) + (\sqrt{3})(\sqrt{2})$

$= 3(\sqrt{5})^2 + 3\sqrt{5 \times 2} + \sqrt{3 \times 5} + \sqrt{3 \times 2}$

$= 3(5) + 3\sqrt{10} + \sqrt{15} + \sqrt{6}$

$= 15 + 3\sqrt{10} + \sqrt{15} + \sqrt{6}$

Denominator: $(\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3$

Combine the numerator and denominator.

$= \frac{15 + 3\sqrt{10} + \sqrt{15} + \sqrt{6}}{3}$

---

(ix) Rationalise the denominator of $\frac{4\sqrt{3} \;+\; 5\sqrt{2}}{\sqrt{48} \;+\; \sqrt{18}}$:

First, simplify the radicals in the denominator.

$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

Substitute these simplified forms into the expression.

$\frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}}$

The denominator is $4\sqrt{3} + 3\sqrt{2}$. Multiply the numerator and the denominator by the conjugate, which is $4\sqrt{3} - 3\sqrt{2}$.

$\frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} = \frac{4\sqrt{3} + 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}} \times \frac{4\sqrt{3} - 3\sqrt{2}}{4\sqrt{3} - 3\sqrt{2}}$

Use $(a+b)(a-b) = a^2 - b^2$ in the denominator.

$= \frac{(4\sqrt{3} + 5\sqrt{2})(4\sqrt{3} - 3\sqrt{2})}{(4\sqrt{3})^2 - (3\sqrt{2})^2}$

Multiply the terms in the numerator (using FOIL or distribution).

Numerator: $(4\sqrt{3})(4\sqrt{3}) - (4\sqrt{3})(3\sqrt{2}) + (5\sqrt{2})(4\sqrt{3}) - (5\sqrt{2})(3\sqrt{2})$

$= 16(\sqrt{3})^2 - 12\sqrt{3 \times 2} + 20\sqrt{2 \times 3} - 15(\sqrt{2})^2$

$= 16(3) - 12\sqrt{6} + 20\sqrt{6} - 15(2)$

$= 48 - 12\sqrt{6} + 20\sqrt{6} - 30$

Combine like terms.

$= (48 - 30) + (-12\sqrt{6} + 20\sqrt{6})$

$= 18 + 8\sqrt{6}$

Denominator: $(4\sqrt{3})^2 - (3\sqrt{2})^2 = (4^2 \times (\sqrt{3})^2) - (3^2 \times (\sqrt{2})^2)$

$= (16 \times 3) - (9 \times 2)$

$= 48 - 18$

$= 30$

Combine the numerator and denominator.

$= \frac{18 + 8\sqrt{6}}{30}$

Factor out the greatest common divisor from the numerator (2) and simplify the fraction.

$= \frac{2(9 + 4\sqrt{6})}{30}$

$= \frac{\cancel{2}^1(9 + 4\sqrt{6})}{\cancel{30}_{15}}$

$= \frac{9 + 4\sqrt{6}}{15}$

---

(i) Find the values of a and b in $\frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}} = a - 6\sqrt{3}$:

Consider the Left Hand Side (LHS): $\frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $7 - 4\sqrt{3}$.

LHS $= \frac{5 \;+\; 2\sqrt{3}}{7 \;+\; 4\sqrt{3}} \times \frac{7 \;-\; 4\sqrt{3}}{7 \;-\; 4\sqrt{3}}$

LHS $= \frac{(5 \;+\; 2\sqrt{3})(7 \;-\; 4\sqrt{3})}{(7 \;+\; 4\sqrt{3})(7 \;-\; 4\sqrt{3})}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (7)^2 - (4\sqrt{3})^2 = 49 - (4^2 \times (\sqrt{3})^2) = 49 - (16 \times 3) = 49 - 48 = 1$.

Using distribution (FOIL) in the numerator:

Numerator $= 5(7) + 5(-4\sqrt{3}) + (2\sqrt{3})(7) + (2\sqrt{3})(-4\sqrt{3})$

Numerator $= 35 - 20\sqrt{3} + 14\sqrt{3} - 8(\sqrt{3})^2$

Numerator $= 35 - 6\sqrt{3} - 8(3)$

Numerator $= 35 - 6\sqrt{3} - 24$

Numerator $= (35 - 24) - 6\sqrt{3} = 11 - 6\sqrt{3}$.

So, LHS $= \frac{11 - 6\sqrt{3}}{1} = 11 - 6\sqrt{3}$.

Now equate LHS to RHS:

Comparing the rational parts on both sides:

Comparing the irrational parts on both sides:

This equation is satisfied and does not involve a parameter $b$ to be found. The structure of the RHS $a - 6\sqrt{3}$ implies that the coefficient of $\sqrt{3}$ is fixed at -6.

Thus, the value of $a$ is 11.

---

(ii) Find the values of a and b in $\frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}} = a\sqrt{5} - \frac{19}{11}$:

Consider the Left Hand Side (LHS): $\frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3 - 2\sqrt{5}$.

LHS $= \frac{3 \;-\; \sqrt{5}}{3 \;+\; 2\sqrt{5}} \times \frac{3 \;-\; 2\sqrt{5}}{3 \;-\; 2\sqrt{5}}$

LHS $= \frac{(3 \;-\; \sqrt{5})(3 \;-\; 2\sqrt{5})}{(3 \;+\; 2\sqrt{5})(3 \;-\; 2\sqrt{5})}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (3)^2 - (2\sqrt{5})^2 = 9 - (4 \times 5) = 9 - 20 = -11$.

Using distribution (FOIL) in the numerator:

Numerator $= 3(3) + 3(-2\sqrt{5}) + (-\sqrt{5})(3) + (-\sqrt{5})(-2\sqrt{5})$

Numerator $= 9 - 6\sqrt{5} - 3\sqrt{5} + 2(\sqrt{5})^2$

Numerator $= 9 - 9\sqrt{5} + 2(5)$

Numerator $= 9 - 9\sqrt{5} + 10$

Numerator $= (9 + 10) - 9\sqrt{5} = 19 - 9\sqrt{5}$.

So, LHS $= \frac{19 - 9\sqrt{5}}{-11} = -\frac{19}{11} + \frac{9\sqrt{5}}{11}$.

Now equate LHS to RHS:

Rearrange the RHS to group rational and irrational parts:

Comparing the rational parts on both sides:

This is an identity and does not help find $a$ or $b$.

Comparing the irrational parts on both sides:

Divide both sides by $\sqrt{5}$ (since $\sqrt{5} \neq 0$):

Thus, the value of $a$ is $\frac{9}{11}$. There is no parameter $b$ in the given equation to find.

---

(iii) Find the values of a and b in $\frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}} = 2 - b\sqrt{6}$:

Consider the Left Hand Side (LHS): $\frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}}$.

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3\sqrt{2} + 2\sqrt{3}$.

LHS $= \frac{\sqrt{2} \;+\; \sqrt{3}}{3\sqrt{2} \;-\; 2\sqrt{3}} \times \frac{3\sqrt{2} \;+\; 2\sqrt{3}}{3\sqrt{2} \;+\; 2\sqrt{3}}$

LHS $= \frac{(\sqrt{2} \;+\; \sqrt{3})(3\sqrt{2} \;+\; 2\sqrt{3})}{(3\sqrt{2} \;-\; 2\sqrt{3})(3\sqrt{2} \;+\; 2\sqrt{3})}$

Using the identity $(x-y)(x+y) = x^2 - y^2$ in the denominator:

Denominator $= (3\sqrt{2})^2 - (2\sqrt{3})^2 = (3^2 \times (\sqrt{2})^2) - (2^2 \times (\sqrt{3})^2) = (9 \times 2) - (4 \times 3) = 18 - 12 = 6$.

Using distribution (FOIL) in the numerator:

Numerator $= (\sqrt{2})(3\sqrt{2}) + (\sqrt{2})(2\sqrt{3}) + (\sqrt{3})(3\sqrt{2}) + (\sqrt{3})(2\sqrt{3})$

Numerator $= 3(\sqrt{2})^2 + 2\sqrt{2 \times 3} + 3\sqrt{3 \times 2} + 2(\sqrt{3})^2$

Numerator $= 3(2) + 2\sqrt{6} + 3\sqrt{6} + 2(3)

Numerator $= 6 + 5\sqrt{6} + 6$

Numerator $= (6 + 6) + 5\sqrt{6} = 12 + 5\sqrt{6}$.

So, LHS $= \frac{12 + 5\sqrt{6}}{6} = \frac{12}{6} + \frac{5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}$.

Now equate LHS to RHS:

Comparing the rational parts on both sides:

This is an identity and does not help find $a$ or $b$.

Comparing the irrational parts on both sides:

Divide both sides by $\sqrt{6}$ (since $\sqrt{6} \neq 0$):

Multiply both sides by -1:

Thus, the value of $b$ is $-\frac{5}{6}$. There is no parameter $a$ in the given equation to find its specific value.

---

(iv) Find the values of a and b in $\frac{(7 \;+\; \sqrt{5})}{(7 \;-\; \sqrt{5})} - \frac{(7 \;-\; \sqrt{5})}{(7 \;+\; \sqrt{5})} = a + \frac{7}{11}\sqrt{5}b$:

Consider the Left Hand Side (LHS): $\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} - \frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}}$.

Rationalize the first term $\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}}$:

$\frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} = \frac{7 \;+\; \sqrt{5}}{7 \;-\; \sqrt{5}} \times \frac{7 \;+\; \sqrt{5}}{7 \;+\; \sqrt{5}} = \frac{(7+\sqrt{5})^2}{7^2 - (\sqrt{5})^2} = \frac{49 + 14\sqrt{5} + 5}{49 - 5} = \frac{54 + 14\sqrt{5}}{44}$

Simplify the fraction by dividing numerator and denominator by 2:

$= \frac{\cancel{54}^{27} + \cancel{14}^{7}\sqrt{5}}{\cancel{44}^{22}} = \frac{27 + 7\sqrt{5}}{22}$.

Rationalize the second term $\frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}}$:

$\frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}} = \frac{7 \;-\; \sqrt{5}}{7 \;+\; \sqrt{5}} \times \frac{7 \;-\; \sqrt{5}}{7 \;-\; \sqrt{5}} = \frac{(7-\sqrt{5})^2}{7^2 - (\sqrt{5})^2} = \frac{49 - 14\sqrt{5} + 5}{49 - 5} = \frac{54 - 14\sqrt{5}}{44}$

Simplify the fraction by dividing numerator and denominator by 2:

$= \frac{\cancel{54}^{27} - \cancel{14}^{7}\sqrt{5}}{\cancel{44}^{22}} = \frac{27 - 7\sqrt{5}}{22}$.

Now subtract the second simplified term from the first simplified term:

LHS $= \frac{27 + 7\sqrt{5}}{22} - \frac{27 - 7\sqrt{5}}{22}$

Since the denominators are the same, subtract the numerators:

LHS $= \frac{(27 + 7\sqrt{5}) - (27 - 7\sqrt{5})}{22}$

LHS $= \frac{27 + 7\sqrt{5} - 27 + 7\sqrt{5}}{22}$

LHS $= \frac{(27 - 27) + (7\sqrt{5} + 7\sqrt{5})}{22}$

LHS $= \frac{0 + 14\sqrt{5}}{22} = \frac{14\sqrt{5}}{22}$

Simplify the fraction by dividing numerator and denominator by 2:

LHS $= \frac{\cancel{14}^{7}\sqrt{5}}{\cancel{22}^{11}} = \frac{7\sqrt{5}}{11}$.

Now equate LHS to RHS:

Rewrite the LHS to clearly show the rational and irrational parts:

Comparing the rational parts on both sides:

Comparing the irrational parts on both sides:

Divide both sides by $\frac{7}{11}\sqrt{5}$ (since $\frac{7}{11}\sqrt{5} \neq 0$):

Thus, the value of $a$ is 0 and the value of $b$ is 1.

---

Given that $a = 2 + \sqrt{3}$.

First, let's find the value of $\frac{1}{a}$.

$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$

To simplify $\frac{1}{2 + \sqrt{3}}$, we rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, which is $2 - \sqrt{3}$.

$\frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

Denominator $= (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

Numerator $= 1 \times (2 - \sqrt{3}) = 2 - \sqrt{3}$.

So, $\frac{1}{a} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$.

Now, we need to find the value of $a - \frac{1}{a}$.

Substitute the values of $a$ and $\frac{1}{a}$ into the expression:

$a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3})$

Remove the parentheses, remembering to change the signs of the terms inside the second parenthesis:

$a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3}$

Combine like terms (rational terms and irrational terms):

$a - \frac{1}{a} = (2 - 2) + (\sqrt{3} + \sqrt{3})$

$a - \frac{1}{a} = 0 + 2\sqrt{3}$

$a - \frac{1}{a} = 2\sqrt{3}$

Thus, the value of $a - \frac{1}{a}$ is $2\sqrt{3}$.

---

(i) Simplify and evaluate $\frac{4}{\sqrt{3}}$:

Rationalise the denominator by multiplying the numerator and denominator by $\sqrt{3}$.

$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{(\sqrt{3})^2} = \frac{4\sqrt{3}}{3}$

Now, substitute the approximate value $\sqrt{3} \approx 1.732$.

$\frac{4\sqrt{3}}{3} \approx \frac{4 \times 1.732}{3}$

Calculate the numerator:

$4 \times 1.732 = 6.928$

Now divide by 3:

$\frac{6.928}{3} \approx 2.30933...$

Rounding to three places of decimal, the value is $2.309$.

---

(ii) Simplify and evaluate $\frac{6}{\sqrt{6}}$:

Rationalise the denominator by multiplying the numerator and denominator by $\sqrt{6}$.

$\frac{6}{\sqrt{6}} = \frac{6}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{6\sqrt{6}}{(\sqrt{6})^2} = \frac{6\sqrt{6}}{6}$

Simplify the fraction:

$\frac{\cancel{6}\sqrt{6}}{\cancel{6}} = \sqrt{6}$

Now, substitute the approximate values $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$.

$\sqrt{6} \approx 1.414 \times 1.732$

Calculate the product:

$1.414 \times 1.732 \approx 2.449048$

Rounding to three places of decimal, the value is $2.449$.

---

(iii) Simplify and evaluate $\frac{\sqrt{10} \;-\; \sqrt{5}}{2}$:

The denominator is already a rational number (2), so no rationalization is needed.

We can rewrite $\sqrt{10}$ as $\sqrt{2 \times 5} = \sqrt{2} \times \sqrt{5}$.

So, the expression is $\frac{\sqrt{2}\sqrt{5} \;-\; \sqrt{5}}{2}$.

Now, substitute the approximate values $\sqrt{2} \approx 1.414$ and $\sqrt{5} \approx 2.236$.

$\frac{\sqrt{10} \;-\; \sqrt{5}}{2} \approx \frac{(1.414 \times 2.236) - 2.236}{2}$

Calculate the term inside the parenthesis:

$1.414 \times 2.236 \approx 3.162224$

Now perform the subtraction in the numerator:

$3.162224 - 2.236 = 0.926224$

Now divide by 2:

$\frac{0.926224}{2} = 0.463112$

Rounding to three places of decimal, the value is $0.463$.

---

(iv) Simplify and evaluate $\frac{\sqrt{2}}{2 \;+\; \sqrt{2}}$:

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $2 - \sqrt{2}$.

$\frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$

$= \frac{\sqrt{2}(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2}$

$= \frac{2\sqrt{2} - (\sqrt{2})^2}{4 - 2}$

$= \frac{2\sqrt{2} - 2}{2}$

Factor out 2 from the numerator and simplify:

$= \frac{\cancel{2}(\sqrt{2} - 1)}{\cancel{2}}$

$= \sqrt{2} - 1$

Now, substitute the approximate value $\sqrt{2} \approx 1.414$.

$\sqrt{2} - 1 \approx 1.414 - 1$

$= 0.414$

The value is already in three decimal places, so the value is $0.414$.

---

(v) Simplify and evaluate $\frac{1}{\sqrt{3} \;+\; \sqrt{2}}$:

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - \sqrt{2}$.

$\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$= \frac{1 \times (\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2}$

$= \frac{\sqrt{3} - \sqrt{2}}{1}$

$= \sqrt{3} - \sqrt{2}$

Now, substitute the approximate values $\sqrt{3} \approx 1.732$ and $\sqrt{2} \approx 1.414$.

$\sqrt{3} - \sqrt{2} \approx 1.732 - 1.414$

$= 0.318$

The value is already in three decimal places, so the value is $0.318$.

---

(i) Simplify $(1^{3} + 2^{3} + 3^{3})^{\frac{1}{2}}$:

Calculate the cubes inside the parenthesis:

$1^3 = 1$

$2^3 = 2 \times 2 \times 2 = 8$

$3^3 = 3 \times 3 \times 3 = 27$

Sum the results:

$1 + 8 + 27 = 36$

Now, evaluate the expression with the sum:

$(36)^{\frac{1}{2}}$

This means finding the square root of 36:

$\sqrt{36} = 6$

So, $(1^{3} + 2^{3} + 3^{3})^{\frac{1}{2}} = 6$.

---

(ii) Simplify $\left( \frac{3}{5} \right)^{4} . \left( \frac{8}{5} \right)^{-12} . \left( \frac{32}{5} \right)^{6}$:

Rewrite the term with the negative exponent using $a^{-n} = \frac{1}{a^n}$ or $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$:

$\left( \frac{8}{5} \right)^{-12} = \left( \frac{5}{8} \right)^{12}$

Express the numbers 8 and 32 as powers of 2: $8 = 2^3$, $32 = 2^5$.

Substitute these into the expression:

$\left( \frac{3}{5} \right)^{4} \times \left( \frac{5}{2^3} \right)^{12} \times \left( \frac{2^5}{5} \right)^{6}$

Apply the exponent rule $(\frac{a}{b})^m = \frac{a^m}{b^m}$ and $(a^m)^n = a^{mn}$:

$= \frac{3^4}{5^4} \times \frac{5^{12}}{(2^3)^{12}} \times \frac{(2^5)^6}{5^6}$

$= \frac{3^4}{5^4} \times \frac{5^{12}}{2^{36}} \times \frac{2^{30}}{5^6}$

Group terms with the same base and use $a^m \times a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$:

$= 3^4 \times \frac{5^{12}}{5^4 \times 5^6} \times \frac{2^{30}}{2^{36}}$

$= 3^4 \times \frac{5^{12}}{5^{4+6}} \times 2^{30-36}$

$= 3^4 \times \frac{5^{12}}{5^{10}} \times 2^{-6}$

$= 3^4 \times 5^{12-10} \times 2^{-6}$

$= 3^4 \times 5^2 \times 2^{-6}$

Rewrite $2^{-6}$ as $\frac{1}{2^6}$:

$= 3^4 \times 5^2 \times \frac{1}{2^6}$

Calculate the powers:

$3^4 = 3 \times 3 \times 3 \times 3 = 81$

$5^2 = 5 \times 5 = 25$

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$

Substitute the values:

$= 81 \times 25 \times \frac{1}{64}$

$= \frac{81 \times 25}{64}$

Calculate the product in the numerator:

$81 \times 25 = 2025$

The simplified expression is $\frac{2025}{64}$.

---

(iii) Simplify $\left( \frac{1}{27} \right)^{-\frac{2}{3}}$:

Rewrite the term with the negative exponent:

$\left( \frac{1}{27} \right)^{-\frac{2}{3}} = \left( \frac{27}{1} \right)^{\frac{2}{3}} = 27^{\frac{2}{3}}$

Express the base 27 as a power of 3: $27 = 3^3$.

Substitute the base:

$(3^3)^{\frac{2}{3}}$

Apply the exponent rule $(a^m)^n = a^{mn}$:

$= 3^{3 \times \frac{2}{3}}$

$= 3^{2}$

Calculate the power:

$3^2 = 9$

So, $\left( \frac{1}{27} \right)^{-\frac{2}{3}} = 9$.

---

(iv) Simplify $\left( \left( \left( 625 \right)^{-\frac{1}{2}} \right)^{-\frac{1}{4}} \right)^{2}$:

Apply the exponent rule $(a^m)^n = a^{mn}$ repeatedly by multiplying the exponents:

The exponents are $-\frac{1}{2}$, $-\frac{1}{4}$, and $2$.

Combined exponent $= (-\frac{1}{2}) \times (-\frac{1}{4}) \times 2$

$= (\frac{1}{8}) \times 2$

$= \frac{2}{8}$

$= \frac{1}{4}$

So, the expression simplifies to:

$625^{\frac{1}{4}}$

This means finding the fourth root of 625.

Express the base 625 as a power of a number. We know $5 \times 5 \times 5 \times 5 = 625$, so $625 = 5^4$.

Substitute the base:

$(5^4)^{\frac{1}{4}}$

Apply the exponent rule $(a^m)^n = a^{mn}$:

$= 5^{4 \times \frac{1}{4}}$

$= 5^{1}$

$= 5$

So, $\left( \left( \left( 625 \right)^{-\frac{1}{2}} \right)^{-\frac{1}{4}} \right)^{2} = 5$.

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(v) Simplify $\frac{9^{\frac{1}{3}} \;\times\; 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \;\times\; 3^{-\frac{2}{3}}}$:

Express all bases as powers of 3: $9 = 3^2$ and $27 = 3^3$.

Substitute these into the expression:

$\frac{(3^2)^{\frac{1}{3}} \;\times\; (3^3)^{-\frac{1}{2}}}{3^{\frac{1}{6}} \;\times\; 3^{-\frac{2}{3}}}$

Apply the exponent rule $(a^m)^n = a^{mn}$ in the numerator:

$= \frac{3^{2 \times \frac{1}{3}} \;\times\; 3^{3 \times (-\frac{1}{2})}}{3^{\frac{1}{6}} \;\times\; 3^{-\frac{2}{3}}}$

$= \frac{3^{\frac{2}{3}} \;\times\; 3^{-\frac{3}{2}}}{3^{\frac{1}{6}} \;\times\; 3^{-\frac{2}{3}}}$

Apply the exponent rule $a^m \times a^n = a^{m+n}$ in the numerator and the denominator:

$= \frac{3^{\frac{2}{3} + (-\frac{3}{2})}}{3^{\frac{1}{6} + (-\frac{2}{3})}}$

Calculate the exponent in the numerator:

$\frac{2}{3} - \frac{3}{2} = \frac{2 \times 2}{3 \times 2} - \frac{3 \times 3}{2 \times 3} = \frac{4}{6} - \frac{9}{6} = \frac{4 - 9}{6} = -\frac{5}{6}$

Calculate the exponent in the denominator:

$\frac{1}{6} - \frac{2}{3} = \frac{1}{6} - \frac{2 \times 2}{3 \times 2} = \frac{1}{6} - \frac{4}{6} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$

Substitute the calculated exponents back into the expression:

$= \frac{3^{-\frac{5}{6}}}{3^{-\frac{1}{2}}}$

Apply the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 3^{-\frac{5}{6} - (-\frac{1}{2})}$

$= 3^{-\frac{5}{6} + \frac{1}{2}}$

Calculate the final exponent:

$-\frac{5}{6} + \frac{1}{2} = -\frac{5}{6} + \frac{1 \times 3}{2 \times 3} = -\frac{5}{6} + \frac{3}{6} = \frac{-5 + 3}{6} = \frac{-2}{6} = -\frac{1}{3}$

The simplified expression is:

$3^{-\frac{1}{3}}$

Rewrite with a positive exponent:

$= \frac{1}{3^{\frac{1}{3}}}$

This can also be written in radical form as $\frac{1}{\sqrt[3]{3}}$.

---

(vi) Simplify $64^{-\frac{1}{3}} . 64^{\frac{1}{3}} - 64^{\frac{2}{3}}$:

Consider the first part $64^{-\frac{1}{3}} \times 64^{\frac{1}{3}}$. Apply the exponent rule $a^m \times a^n = a^{m+n}$:

$64^{-\frac{1}{3}} \times 64^{\frac{1}{3}} = 64^{-\frac{1}{3} + \frac{1}{3}} = 64^0$

Any non-zero number raised to the power of 0 is 1:

$64^0 = 1$

So the expression becomes:

$1 - 64^{\frac{2}{3}}$

Calculate $64^{\frac{2}{3}}$. This can be written as $(64^{\frac{1}{3}})^2$.

First, find the cube root of 64: $\sqrt[3]{64}$. We know that $4 \times 4 \times 4 = 64$, so $\sqrt[3]{64} = 4$.

Now, square the result:

$(4)^2 = 16$

Substitute this value back into the expression:

$1 - 16$

$= -15$

So, $64^{-\frac{1}{3}} . 64^{\frac{1}{3}} - 64^{\frac{2}{3}} = -15$.

---

(vii) Simplify $\frac{8^{\frac{1}{3}} \;\times\; 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$:

Express all bases as powers of 2: $8 = 2^3$, $16 = 2^4$, $32 = 2^5$.

Substitute these into the expression:

$\frac{(2^3)^{\frac{1}{3}} \;\times\; (2^4)^{\frac{1}{3}}}{(2^5)^{-\frac{1}{3}}}$

Apply the exponent rule $(a^m)^n = a^{mn}$:

$= \frac{2^{3 \times \frac{1}{3}} \;\times\; 2^{4 \times \frac{1}{3}}}{2^{5 \times (-\frac{1}{3})}}$

$= \frac{2^{1} \;\times\; 2^{\frac{4}{3}}}{2^{-\frac{5}{3}}}$

Apply the exponent rule $a^m \times a^n = a^{m+n}$ in the numerator:

$= \frac{2^{1 + \frac{4}{3}}}{2^{-\frac{5}{3}}}$

Calculate the exponent in the numerator:

$1 + \frac{4}{3} = \frac{3}{3} + \frac{4}{3} = \frac{3 + 4}{3} = \frac{7}{3}$

The expression becomes:

$= \frac{2^{\frac{7}{3}}}{2^{-\frac{5}{3}}}$

Apply the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$= 2^{\frac{7}{3} - (-\frac{5}{3})}$

$= 2^{\frac{7}{3} + \frac{5}{3}}$

Calculate the final exponent:

$= 2^{\frac{7 + 5}{3}}$

$= 2^{\frac{12}{3}}$

$= 2^4$

Calculate the power:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$

So, $\frac{8^{\frac{1}{3}} \;\times\; 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}} = 16$.

Given:

$a = 5 + 2\sqrt{6}$

$b = \frac{1}{a}$

To Find:

The value of $a^2 + b^2$.

---

Solution:

We are given $a = 5 + 2\sqrt{6}$.

We are also given that $b = \frac{1}{a}$. Let's find the value of $b$ by substituting the value of $a$:

$b = \frac{1}{5 + 2\sqrt{6}}$

To simplify this expression and rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $5 - 2\sqrt{6}$.

$b = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}}$

In the denominator, we use the difference of squares formula: $(x+y)(x-y) = x^2 - y^2$. Here, $x=5$ and $y=2\sqrt{6}$.

$b = \frac{5 - 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2}$

$b = \frac{5 - 2\sqrt{6}}{25 - (2^2 \times (\sqrt{6})^2)}$

$b = \frac{5 - 2\sqrt{6}}{25 - (4 \times 6)}$

$b = \frac{5 - 2\sqrt{6}}{25 - 24}$

$b = \frac{5 - 2\sqrt{6}}{1}$

$b = 5 - 2\sqrt{6}$

So, we have $a = 5 + 2\sqrt{6}$ and $b = 5 - 2\sqrt{6}$.

We need to find the value of $a^2 + b^2$. We can use the algebraic identity $a^2 + b^2 = (a+b)^2 - 2ab$.

First, let's find the sum $a+b$:

$a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$

$a + b = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}$

$a + b = (5 + 5) + (2\sqrt{6} - 2\sqrt{6})$

$a + b = 10 + 0$

$a + b = 10$

Next, let's find the product $ab$:

$ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6})$

Using the identity $(x+y)(x-y) = x^2 - y^2$ again:

$ab = (5)^2 - (2\sqrt{6})^2$

$ab = 25 - (4 \times 6)$

$ab = 25 - 24$

$ab = 1$

Now, substitute the values of $(a+b)$ and $ab$ into the identity $a^2 + b^2 = (a+b)^2 - 2ab$:

$a^2 + b^2 = (10)^2 - 2(1)$

$a^2 + b^2 = 100 - 2$

$a^2 + b^2 = 98$

---

Alternate Solution:

We can also calculate $a^2$ and $b^2$ separately and then add them.

Given $a = 5 + 2\sqrt{6}$. Calculate $a^2$ using the identity $(x+y)^2 = x^2 + 2xy + y^2$:

$a^2 = (5 + 2\sqrt{6})^2$

$a^2 = (5)^2 + 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$a^2 = 25 + 20\sqrt{6} + (4 \times 6)$

$a^2 = 25 + 20\sqrt{6} + 24$

$a^2 = 49 + 20\sqrt{6}$

From the primary solution, we found $b = 5 - 2\sqrt{6}$. Calculate $b^2$ using the identity $(x-y)^2 = x^2 - 2xy + y^2$:

$b^2 = (5 - 2\sqrt{6})^2$

$b^2 = (5)^2 - 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$b^2 = 25 - 20\sqrt{6} + (4 \times 6)$

$b^2 = 25 - 20\sqrt{6} + 24$

$b^2 = 49 - 20\sqrt{6}$

Now, add $a^2$ and $b^2$:

$a^2 + b^2 = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6})$

$a^2 + b^2 = 49 + 20\sqrt{6} + 49 - 20\sqrt{6}$

$a^2 + b^2 = (49 + 49) + (20\sqrt{6} - 20\sqrt{6})$

$a^2 + b^2 = 98 + 0$

$a^2 + b^2 = 98$

---

The value of $a^2 + b^2$ is 98.

---

We need to express each term in the sum in the form $\frac{p}{q}$.

Term 1: 0.6

$0.6 = \frac{6}{10}$

Simplify the fraction:

$0.6 = \frac{\cancel{6}^3}{\cancel{10}_5} = \frac{3}{5}$

---

Term 2: 0.$\overline{7}$

Let $x = 0.\overline{7}$

Multiply equation (i) by 10 (since one digit is repeating):

Subtract equation (i) from equation (ii):

Solve for $x$:

So, $0.\overline{7} = \frac{7}{9}$.

---

Term 3: 0.4$\overline{7}$

Let $y = 0.4\overline{7}$

Multiply equation (iii) by 10 (to move the non-repeating digit past the decimal):

Multiply equation (iv) by 10 (since one digit is repeating after the decimal in (iv)):

Subtract equation (iv) from equation (v):

Solve for $y$:

So, $0.4\overline{7} = \frac{43}{90}$.

---

Adding the fractions:

Now we need to calculate the sum: $0.6 + 0.\overline{7} + 0.4\overline{7} = \frac{3}{5} + \frac{7}{9} + \frac{43}{90}$.

Find the Least Common Multiple (LCM) of the denominators 5, 9, and 90.

Prime factorization of the denominators:

$5 = 5^1$

$9 = 3^2$

$90 = 2 \times 45 = 2 \times 9 \times 5 = 2^1 \times 3^2 \times 5^1$

LCM$(5, 9, 90) = 2^1 \times 3^2 \times 5^1 = 2 \times 9 \times 5 = 90$.

Convert each fraction to have a denominator of 90:

$\frac{3}{5} = \frac{3 \times 18}{5 \times 18} = \frac{54}{90}$

$\frac{7}{9} = \frac{7 \times 10}{9 \times 10} = \frac{70}{90}$

$\frac{43}{90}$ (already has the denominator 90)

Add the fractions:

$\frac{54}{90} + \frac{70}{90} + \frac{43}{90} = \frac{54 + 70 + 43}{90}$

Calculate the sum in the numerator:

$54 + 70 = 124$

$124 + 43 = 167$

The sum is $\frac{167}{90}$.

The fraction $\frac{167}{90}$ is in the form $\frac{p}{q}$, where $p=167$ and $q=90$. Both are integers and $q \neq 0$. The fraction is already in its simplest form as 167 is a prime number and 90 is not divisible by 167.

Thus, $0.6 + 0.\overline{7} + 0.4\overline{7} = \frac{167}{90}$.

---

We need to simplify the given expression by rationalizing the denominator of each term.

Term 1: $\frac{7\sqrt{3}}{\sqrt{10} \;+\; \sqrt{3}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{10} - \sqrt{3}$.

$\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} \times \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} = \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10})^2 - (\sqrt{3})^2}$

$= \frac{7\sqrt{30} - 7(\sqrt{3})^2}{10 - 3} = \frac{7\sqrt{30} - 7(3)}{7} = \frac{7\sqrt{30} - 21}{7}$

$= \frac{7(\sqrt{30} - 3)}{7} = \sqrt{30} - 3$

---

Term 2: $\frac{2\sqrt{5}}{\sqrt{6} \;+\; \sqrt{5}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{6} - \sqrt{5}$.

$\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{(\sqrt{6})^2 - (\sqrt{5})^2}$

$= \frac{2\sqrt{30} - 2(\sqrt{5})^2}{6 - 5} = \frac{2\sqrt{30} - 2(5)}{1} = 2\sqrt{30} - 10$

---

Term 3: $\frac{3\sqrt{2}}{\sqrt{15} \;+\; 3\sqrt{2}}$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{15} - 3\sqrt{2}$.

$\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \times \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} = \frac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{(\sqrt{15})^2 - (3\sqrt{2})^2}$

$= \frac{3\sqrt{30} - 3(3\sqrt{2})(\sqrt{2})}{15 - (3^2 \times (\sqrt{2})^2)} = \frac{3\sqrt{30} - 9(2)}{15 - (9 \times 2)} = \frac{3\sqrt{30} - 18}{15 - 18}$

$= \frac{3\sqrt{30} - 18}{-3} = \frac{3(\sqrt{30} - 6)}{-3} = -(\sqrt{30} - 6) = -\sqrt{30} + 6$

---

Now, substitute the simplified terms back into the original expression:

$(\sqrt{30} - 3) - (2\sqrt{30} - 10) - (-\sqrt{30} + 6)$

Remove the parentheses, being careful with the signs:

$= \sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6$

Group the terms with $\sqrt{30}$ and the constant terms:

$= (\sqrt{30} - 2\sqrt{30} + \sqrt{30}) + (-3 + 10 - 6)$

Combine the coefficients of $\sqrt{30}$:

$= (1 - 2 + 1)\sqrt{30} + (-3 + 10 - 6)$

$= 0\sqrt{30} + (7 - 6)$

$= 0 + 1$

$= 1$

The simplified value of the expression is 1.

Given:

Expression: $\frac{4}{(3\sqrt{3} \;-\; 2\sqrt{2})}+\frac{3}{(3\sqrt{3} \;+\; 2\sqrt{2})}$

Approximate values: $\sqrt{2} = 1.414$, $\sqrt{3} = 1.732$

To Find:

The value of the expression using the given approximations, rounded to three decimal places.

---

Solution:

First, we simplify the given expression by combining the two fractions. The denominators are conjugates of each other, so we can find a common denominator by multiplying them.

Common denominator = $(3\sqrt{3} - 2\sqrt{2})(3\sqrt{3} + 2\sqrt{2})$

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$(3\sqrt{3})^2 - (2\sqrt{2})^2 = (3^2 \times (\sqrt{3})^2) - (2^2 \times (\sqrt{2})^2)$

$= (9 \times 3) - (4 \times 2) = 27 - 8 = 19$

Now, combine the fractions over the common denominator:

$\frac{4(3\sqrt{3} \;+\; 2\sqrt{2}) \;+\; 3(3\sqrt{3} \;-\; 2\sqrt{2})}{(3\sqrt{3} \;-\; 2\sqrt{2})(3\sqrt{3} \;+\; 2\sqrt{2})}$

$= \frac{4(3\sqrt{3}) \;+\; 4(2\sqrt{2}) \;+\; 3(3\sqrt{3}) \;-\; 3(2\sqrt{2})}{19}$

$= \frac{12\sqrt{3} \;+\; 8\sqrt{2} \;+\; 9\sqrt{3} \;-\; 6\sqrt{2}}{19}$

Group like terms in the numerator (terms with $\sqrt{3}$ and terms with $\sqrt{2}$):

$= \frac{(12\sqrt{3} \;+\; 9\sqrt{3}) \;+\; (8\sqrt{2} \;-\; 6\sqrt{2})}{19}$

$= \frac{(12 + 9)\sqrt{3} \;+\; (8 - 6)\sqrt{2}}{19}$

$= \frac{21\sqrt{3} \;+\; 2\sqrt{2}}{19}$

---

Now, substitute the given approximate values $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$ into the simplified expression.

Value $\approx \frac{21(1.732) \;+\; 2(1.414)}{19}$

Calculate the products in the numerator:

$21 \times 1.732 = 36.372$

$2 \times 1.414 = 2.828$

Add these values:

$36.372 + 2.828 = 39.200$

Now, divide the sum by 19:

Value $\approx \frac{39.200}{19}$

Perform the division:

$39.200 \div 19 \approx 2.06315...$

Rounding the result to three places of decimal:

Value $\approx 2.063$

The value of the expression, rounded to three decimal places, is $2.063$.

Given:

$a = \frac{3 \;+\; \sqrt{5}}{2}$

To Find:

The value of $a^2 + \frac{1}{a^2}$.

---

Solution:

We are given the value of $a$. To find $a^2 + \frac{1}{a^2}$, we can first find the value of $\frac{1}{a}$.

$\frac{1}{a} = \frac{1}{\frac{3 \;+\; \sqrt{5}}{2}}$

$\frac{1}{a} = \frac{2}{3 \;+\; \sqrt{5}}$

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator, which is $3 - \sqrt{5}$.

$\frac{1}{a} = \frac{2}{3 \;+\; \sqrt{5}} \times \frac{3 \;-\; \sqrt{5}}{3 \;-\; \sqrt{5}}$

Using the identity $(x+y)(x-y) = x^2 - y^2$ in the denominator:

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{3^2 \;-\; (\sqrt{5})^2}$

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{9 \;-\; 5}$

$\frac{1}{a} = \frac{2(3 \;-\; \sqrt{5})}{4}$

Simplify the fraction:

$\frac{1}{a} = \frac{\cancel{2}^1(3 \;-\; \sqrt{5})}{\cancel{4}_2}$

$\frac{1}{a} = \frac{3 \;-\; \sqrt{5}}{2}$

Now we have $a = \frac{3 + \sqrt{5}}{2}$ and $\frac{1}{a} = \frac{3 - \sqrt{5}}{2}$.

We can find $a^2 + \frac{1}{a^2}$ using the identity $a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$.

First, calculate the sum $a + \frac{1}{a}$:

$a + \frac{1}{a} = \frac{3 \;+\; \sqrt{5}}{2} + \frac{3 \;-\; \sqrt{5}}{2}$

Since the denominators are the same, add the numerators:

$a + \frac{1}{a} = \frac{(3 \;+\; \sqrt{5}) \;+\; (3 \;-\; \sqrt{5})}{2}$

$a + \frac{1}{a} = \frac{3 \;+\; \sqrt{5} \;+\; 3 \;-\; \sqrt{5}}{2}$

Combine the terms in the numerator:

$a + \frac{1}{a} = \frac{(3 \;+\; 3) \;+\; (\sqrt{5} \;-\; \sqrt{5})}{2}$

$a + \frac{1}{a} = \frac{6 \;+\; 0}{2}$

$a + \frac{1}{a} = \frac{6}{2}$

$a + \frac{1}{a} = 3$

Now, substitute the value of $a + \frac{1}{a}$ into the identity $a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$:

$a^2 + \frac{1}{a^2} = (3)^2 - 2$

$a^2 + \frac{1}{a^2} = 9 - 2$

$a^2 + \frac{1}{a^2} = 7$

---

Alternate Solution:

We can directly calculate $a^2$ and $\frac{1}{a^2}$ and then add them.

$a = \frac{3 \;+\; \sqrt{5}}{2}$

$a^2 = \left(\frac{3 \;+\; \sqrt{5}}{2}\right)^2 = \frac{(3 \;+\; \sqrt{5})^2}{2^2}$

$a^2 = \frac{3^2 \;+\; 2(3)(\sqrt{5}) \;+\; (\sqrt{5})^2}{4}$

$a^2 = \frac{9 \;+\; 6\sqrt{5} \;+\; 5}{4}$

$a^2 = \frac{14 \;+\; 6\sqrt{5}}{4}$

Simplify:

$a^2 = \frac{\cancel{2}(7 \;+\; 3\sqrt{5})}{\cancel{4}_2}$

$a^2 = \frac{7 \;+\; 3\sqrt{5}}{2}$

From the first method, we found $\frac{1}{a} = \frac{3 \;-\; \sqrt{5}}{2}$.

So, $\frac{1}{a^2} = \left(\frac{1}{a}\right)^2 = \left(\frac{3 \;-\; \sqrt{5}}{2}\right)^2 = \frac{(3 \;-\; \sqrt{5})^2}{2^2}$

$\frac{1}{a^2} = \frac{3^2 \;-\; 2(3)(\sqrt{5}) \;+\; (\sqrt{5})^2}{4}$

$\frac{1}{a^2} = \frac{9 \;-\; 6\sqrt{5} \;+\; 5}{4}$

$\frac{1}{a^2} = \frac{14 \;-\; 6\sqrt{5}}{4}$

Simplify:

$\frac{1}{a^2} = \frac{\cancel{2}(7 \;-\; 3\sqrt{5})}{\cancel{4}_2}$

$\frac{1}{a^2} = \frac{7 \;-\; 3\sqrt{5}}{2}$

Now, add $a^2$ and $\frac{1}{a^2}$:

$a^2 + \frac{1}{a^2} = \frac{7 \;+\; 3\sqrt{5}}{2} + \frac{7 \;-\; 3\sqrt{5}}{2}$

$a^2 + \frac{1}{a^2} = \frac{(7 \;+\; 3\sqrt{5}) \;+\; (7 \;-\; 3\sqrt{5})}{2}$

$a^2 + \frac{1}{a^2} = \frac{7 \;+\; 3\sqrt{5} \;+\; 7 \;-\; 3\sqrt{5}}{2}$

$a^2 + \frac{1}{a^2} = \frac{(7 \;+\; 7) \;+\; (3\sqrt{5} \;-\; 3\sqrt{5})}{2}$

$a^2 + \frac{1}{a^2} = \frac{14 \;+\; 0}{2}$

$a^2 + \frac{1}{a^2} = \frac{14}{2}$

$a^2 + \frac{1}{a^2} = 7$

The value of $a^2 + \frac{1}{a^2}$ is 7.

Given:

$x = \frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$

$y = \frac{\sqrt{3} \;-\; \sqrt{2}}{\sqrt{3} \;+\; \sqrt{2}}$

To Find:

The value of $x^2 + y^2$.

---

Solution:

First, let's simplify the expressions for $x$ and $y$ by rationalizing their denominators.

For $x = \frac{\sqrt{3} \;+\; \sqrt{2}}{\sqrt{3} \;-\; \sqrt{2}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + \sqrt{2}$.

$x = \frac{(\sqrt{3} \;+\; \sqrt{2})}{(\sqrt{3} \;-\; \sqrt{2})} \times \frac{(\sqrt{3} \;+\; \sqrt{2})}{(\sqrt{3} \;+\; \sqrt{2})}$

$x = \frac{(\sqrt{3} \;+\; \sqrt{2})^2}{(\sqrt{3})^2 \;-\; (\sqrt{2})^2}$

Using $(a+b)^2 = a^2 + 2ab + b^2$ in the numerator and $(a-b)(a+b) = a^2 - b^2$ in the denominator:

$x = \frac{(\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2}{3 - 2}$

$x = \frac{3 + 2\sqrt{6} + 2}{1}$

$x = 5 + 2\sqrt{6}$

For $y = \frac{\sqrt{3} \;-\; \sqrt{2}}{\sqrt{3} \;+\; \sqrt{2}}$, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - \sqrt{2}$.

$y = \frac{(\sqrt{3} \;-\; \sqrt{2})}{(\sqrt{3} \;+\; \sqrt{2})} \times \frac{(\sqrt{3} \;-\; \sqrt{2})}{(\sqrt{3} \;-\; \sqrt{2})}$

$y = \frac{(\sqrt{3} \;-\; \sqrt{2})^2}{(\sqrt{3})^2 \;-\; (\sqrt{2})^2}$

Using $(a-b)^2 = a^2 - 2ab + b^2$ in the numerator and $(a+b)(a-b) = a^2 - b^2$ in the denominator:

$y = \frac{(\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2}{3 - 2}$

$y = \frac{3 - 2\sqrt{6} + 2}{1}$

$y = 5 - 2\sqrt{6}$

So, we have $x = 5 + 2\sqrt{6}$ and $y = 5 - 2\sqrt{6}$.

We can find $x^2 + y^2$ using the algebraic identity $x^2 + y^2 = (x+y)^2 - 2xy$.

First, calculate the sum $x+y$:

$x + y = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6})$

$x + y = 5 + 2\sqrt{6} + 5 - 2\sqrt{6}$

$x + y = (5+5) + (2\sqrt{6} - 2\sqrt{6})$

$x + y = 10 + 0 = 10$

Next, calculate the product $xy$:

$xy = (5 + 2\sqrt{6})(5 - 2\sqrt{6})$

Using the identity $(a+b)(a-b) = a^2 - b^2$:

$xy = (5)^2 - (2\sqrt{6})^2$

$xy = 25 - (4 \times 6)$

$xy = 25 - 24$

$xy = 1$

Now, substitute the values of $(x+y)$ and $xy$ into the identity $x^2 + y^2 = (x+y)^2 - 2xy$:

$x^2 + y^2 = (10)^2 - 2(1)$

$x^2 + y^2 = 100 - 2$

$x^2 + y^2 = 98$

---

Alternate Solution:

We can also calculate $x^2$ and $y^2$ directly and then add them.

We have $x = 5 + 2\sqrt{6}$.

$x^2 = (5 + 2\sqrt{6})^2$

Using $(a+b)^2 = a^2 + 2ab + b^2$:

$x^2 = 5^2 + 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$x^2 = 25 + 20\sqrt{6} + (4 \times 6)$

$x^2 = 25 + 20\sqrt{6} + 24$

$x^2 = 49 + 20\sqrt{6}$

We have $y = 5 - 2\sqrt{6}$.

$y^2 = (5 - 2\sqrt{6})^2$

Using $(a-b)^2 = a^2 - 2ab + b^2$:

$y^2 = 5^2 - 2(5)(2\sqrt{6}) + (2\sqrt{6})^2$

$y^2 = 25 - 20\sqrt{6} + (4 \times 6)$

$y^2 = 25 - 20\sqrt{6} + 24$

$y^2 = 49 - 20\sqrt{6}$

Now, add $x^2$ and $y^2$:

$x^2 + y^2 = (49 + 20\sqrt{6}) + (49 - 20\sqrt{6})$

$x^2 + y^2 = 49 + 20\sqrt{6} + 49 - 20\sqrt{6}$

$x^2 + y^2 = (49 + 49) + (20\sqrt{6} - 20\sqrt{6})$

$x^2 + y^2 = 98 + 0$

$x^2 + y^2 = 98$

The value of $x^2 + y^2$ is 98.

---

We need to simplify the expression $(256)^{(-4)^{\left( \frac{-3}{2} \right)}}$. This is a nested exponent expression, which we simplify from the top exponent downwards.

Let's evaluate the top exponent first: $(-4)^{\left( \frac{-3}{2} \right)}$.

The exponent is $-\frac{3}{2}$. The base is $-4$. The fractional exponent $\frac{3}{2}$ means taking the square root (denominator 2) and then cubing the result (numerator 3).

$(-4)^{-\frac{3}{2}} = \frac{1}{(-4)^{\frac{3}{2}}}$

$= \frac{1}{(\sqrt{-4})^3}$

The term $\sqrt{-4}$ involves the square root of a negative number, which is not a real number ($\sqrt{-4} = 2i$, where $i = \sqrt{-1}$). This would lead to a complex number result for the exponent and consequently for the entire expression.

In the context of typical high school mathematics where such simplification problems result in real numbers, it is highly likely that the base of the inner exponent was intended to be positive 4, i.e., the expression is $(256)^{(4)^{\left( \frac{-3}{2} \right)}}$. We will proceed with this assumption to obtain a real number simplification.

Let's evaluate $(4)^{\left( \frac{-3}{2} \right)}$ assuming the base is 4.

$4^{-\frac{3}{2}} = \frac{1}{4^{\frac{3}{2}}}

$= \frac{1}{(\sqrt{4})^3}$

$= \frac{1}{(2)^3}

$= \frac{1}{8}$

Now substitute this value back into the main expression, assuming the intended form $(256)^{(4)^{-3/2}}$:

$(256)^{\frac{1}{8}}$

This means finding the 8th root of 256. We can express 256 as a power of 2:

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8$

Substitute this into the expression:

$(2^8)^{\frac{1}{8}}$

Apply the exponent rule $(a^m)^n = a^{mn}$:

$= 2^{8 \times \frac{1}{8}}$

$= 2^{1}$

$= 2$

Under the assumption that the inner base is 4, the simplified value of the expression is 2.

(Note: A strict interpretation of the question as written, with base -4 for the inner exponent, results in a complex number.)

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We need to simplify the given expression by evaluating each term.

Recall the exponent rule $\frac{1}{a^{-n}} = a^n$. Using this, we can rewrite the expression as:

$4 \times (216)^{\frac{2}{3}} + 1 \times (256)^{\frac{3}{4}} + 2 \times (216)^{\frac{1}{5}}$

Let's evaluate each term separately.

Term 1: $4 \times (216)^{\frac{2}{3}}$

We can write $216$ as a power of 6, since $6^3 = 216$.

$(216)^{\frac{2}{3}} = (6^3)^{\frac{2}{3}}$

Using the exponent rule $(a^m)^n = a^{mn}$:

$= 6^{3 \times \frac{2}{3}} = 6^2 = 36$

So, the first term is $4 \times 36 = 144$.

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Term 2: $1 \times (256)^{\frac{3}{4}}$

We need to find the fourth root of 256 and then cube it. We know that $4^4 = 256$ and also $2^8 = 256$. Let's use $4^4$ for simplicity.

$(256)^{\frac{3}{4}} = (4^4)^{\frac{3}{4}}$

Using the exponent rule $(a^m)^n = a^{mn}$:

$= 4^{4 \times \frac{3}{4}} = 4^3 = 4 \times 4 \times 4 = 64$

So, the second term is $1 \times 64 = 64$.

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Term 3: $2 \times (216)^{\frac{1}{5}}$

We need to find the fifth root of 216 and then multiply by 2. We know $216 = 6^3$.

$(216)^{\frac{1}{5}} = (6^3)^{\frac{1}{5}}$

Using the exponent rule $(a^m)^n = a^{mn}$:

$= 6^{3 \times \frac{1}{5}} = 6^{\frac{3}{5}}$

This term cannot be simplified to a rational number.

So, the third term is $2 \times 6^{\frac{3}{5}}$.

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Now, add the values of the three terms:

$144 + 64 + 2 \times 6^{\frac{3}{5}}$

Combine the rational numbers:

$144 + 64 = 208$

The value of the expression is $208 + 2 \times 6^{\frac{3}{5}}$.

(Note: If the third term was intended to have a rational value, there might be a typo in the base (e.g., 32 or 243 instead of 216) or the exponent (e.g., -1/3 instead of -1/5). However, based on the provided text, the simplification is $208 + 2 \times 6^{\frac{3}{5}}$).